WBCS Exam Main Compulsory Question Paper on Arithmetic and Test of Reasoning 2016 solved paper
1. How many cubes of 10 cm edge can be put in a cubic box of 1 m edge ?
(1) 10 (2) 100 (3) 1000 (4) 10000
Ans:- The volume of small cube of 10 cm edge is = [tex]{10^3} = 1000[/tex] [tex]c{m^3}[/tex]
The volume of 1 m = 100 cm cubic box is = [tex]{100^3} = 1000000[/tex] [tex]c{m^3}[/tex]
Number of cubes are [tex]\frac{{1000000}}{{1000}} = 1000[/tex]
2. Next terms of the series 198, 194, 185, 169 .....
(1) 92 (2)112 (3)136 (4)144
Ans: The different between 1st and 2nd term is ( 198 - 194 ) = 4=[tex]{2^2}[/tex]
The different between 2nd and 3rd term is ( 194 - 185 ) = 9 = [tex]{3^2}[/tex]
The different between 3rd and 4th term is ( 185 - 169 ) = 16 = [tex]{4^2}[/tex]
Then we can say the different between 4th and 5th term is [tex]{5^2} = 25[/tex]
Then the next term is 169 - 25 = 144
3. Which fraction comes next in the sequence
[tex]\frac{1}{2},\frac{3}{4},\frac{5}{8},\frac{7}{{16}}[/tex]
(1) [tex]\frac{9}{{32}}[/tex] (2) [tex]\frac{{10}}{{17}}[/tex] (3) [tex]\frac{{11}}{{34}}[/tex] (4) [tex]\frac{{12}}{{35}}[/tex]
Ans: The sequence is [tex]\frac{1}{2},\frac{3}{4},\frac{5}{8},\frac{7}{{16}}[/tex]
Then
[tex]\begin{array}{l} \frac{1}{2},\frac{3}{4},\frac{5}{8},\frac{7}{{16}}\\ \frac{1}{{{2^1}}},\frac{{1 + 2}}{{{2^2}}},\frac{{1 + 2 + 2}}{{{2^3}}},\frac{{1 + 2 + 2 + 2}}{{{2^4}}} \end{array}[/tex]
Now the next term will be [tex]\frac{{1 + 2 + 2 + 2 + 2}}{{{2^5}}} = \frac{9}{{32}}[/tex]
4. The last day of century cannot be
(1) Monday (2) Wednesday (3) Tuesday (4) Friday
Ans: 100 years contain 5 odd days. So last day of 1st century is Friday.
200 years contain [tex](5 \times 2) \equiv 3[/tex] odd days. So last day of 2nd century is Wednesday .
300 years contain [tex](5 \times 3) = 15 \equiv 1[/tex] odd day. So last day of 3rd century is Monday.
400 years contain 0 odd day. So the last day of 4th century is Sunday.
this cycle is repeated. So last day of century cannot be Tuesday, Thursday or Saturday.
5. The area of a square is equal to the area of a circle. The ratio between the side of a square and the radius of the circle is
(1) [tex]\sqrt \pi :1[/tex] (2) [tex]1:\sqrt \pi [/tex] (3) [tex]1:\pi [/tex] (4) [tex]\pi :1[/tex]
Ans: Let the one side of the square is a. Then the area of the square is [tex]{a^2}[/tex].
Let the radius of the circle is r. Then the area of the circle is [tex]\pi {r^2}[/tex].
Now we can write
[tex]\begin{array}{l} {a^2} = \pi {r^2}\\ \Rightarrow a = \sqrt \pi r\\ \Rightarrow \frac{a}{r} = \sqrt \pi \\ \Rightarrow a:r = \sqrt \pi :1 \end{array}[/tex]
6. Two trains, each 100 m long moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast as the other, then the speed of the faster train is
(1) 40 km/hr (2) 50 km/hr (3) 60 km/hr (4) 70 km/hr
Ans. The speed of the slower train is u m/sec, then the speed of the faster train is 2u m/sec.
The trains moving in opposite direction,then the relative velocity is (u + 2u ) m/sec = 3u m/sec
Then ( 100 + 100 ) m = 200 m cross time of each train is [tex]\frac{{200}}{{3u}}[/tex] sec.
Now we can write [tex]8 = \frac{{200}}{{3u}} \Rightarrow u = \frac{{200}}{{3 \times 8}} \Rightarrow u = \frac{{25}}{3}[/tex] m/sec
Then the speed of the faster train is [tex]2u = 2 \times \frac{{25}}{3}[/tex] m/sec or [tex]2 \times \frac{{25}}{3} \times \frac{{3600}}{{1000}} = 60[/tex] km/hr.
7. In a river a man takes 3 hour in rowing 3 km up-stream or 15 km down-stream, the speed of the current is
(1) 2 km/hr (2) 4 km/hr (3) 6 km/hr (4) 9 km/hr
Ans. Let the speed of the current is u km/hr and the speed of the man is v km/hr in no stream.
Then the man rowing ( v - u ) km in 1 hour in up-stream and ( v + u ) km in 1 hour in down-stream.
Then we can write 3( v - u ) = 3 and 3( v + u ) =15
[tex]\begin{array}{l}
3(v - u) = 3 \Rightarrow v - u = 1............(1)\\
3(v + u) = 15 \Rightarrow v + u = 5...........(2)\\
(2) - (1)\\
2u = 4 \Rightarrow u = 2
\end{array}[/tex]
The speed of the current is 2 km/hr.
8. In what ratio the water be mixed with milk to gain [tex]16\frac{2}{3}\% [/tex] on selling the mixture at cost price?
(1) 1 : 6 (2) 2 : 3 (3) 4 : 3 (4) 6 : 1
Ans. Let the quantity of the milk is x liters and the amount of the water is y liters.Then C.P of x liters =S.P of ( x + y ) liters.
Gain = [tex]16\frac{2}{3}\% = \frac{{50}}{3}\% [/tex]
[tex]\begin{array}{l}
S.P = C.P(1 + \frac{{Gain\% }}{{100}})\\
\Rightarrow x + y = x(1 + \frac{{50}}{{300}})\\
\Rightarrow x + y = x\frac{{35}}{{30}}\\
\Rightarrow 30x + 30y = 35x\\
\Rightarrow 30y = 5x\\
\Rightarrow \frac{y}{x} = \frac{1}{6}\\
\Rightarrow y:x = 1:6
\end{array}[/tex]
9. Simple interest of Rs. 16,250 at 8% per annum for 73 days is
(1) Rs. 460 (2) Rs. 260 (3) Rs. 560 (3) Rs.660
Ans. We know [tex]S.I = \frac{{P \times R \times T}}{{100}}[/tex]
Here P = Rs.16,250 R = 8% T = [tex]\frac{{73}}{{365}}[/tex] years
Then [tex]S.I = \frac{{16250 \times 8 \times \frac{{73}}{{365}}}}{{100}} = 260[/tex]
S.I = Rs. 260
10. The diagonals of two squares are in the ratio 5 : 2 of their area is
(1) 5 : 2 (2) 25 : 4 (3) 125 : 8 (4) 4 : 25
Ans. Let the sides of two squares are x and y
Then the diagonals of them are [tex](\sqrt {{x^2} + {x^2}} = \sqrt 2 x),(\sqrt {{y^2} + {y^2}} = \sqrt 2 y)[/tex]
Then we can write [tex]\frac{{\sqrt 2 x}}{{\sqrt 2 y}} = \frac{5}{2} \Rightarrow \frac{x}{y} = \frac{5}{2} \Rightarrow x = \frac{5}{2}y[/tex]
The area of them are [tex]{x^2}[/tex] and [tex]{y^2}[/tex]
Then [tex]\frac{{{x^2}}}{{{y^2}}} = \frac{{{{(\frac{5}{2})}^2}{y^2}}}{{{y^2}}} = \frac{{25}}{4}[/tex]
The ratio of the areas is 25 : 4
11. Each side of a rhombus is 5 cm. Its area is
(1) 25 [tex]c{m^2}[/tex] (2) 23 [tex]c{m^2}[/tex] (3) 24 [tex]c{m^2}[/tex] (4) data inadequate
Ans. We know the area of rhombus is [tex]\frac{1}{2} \times {d_1} \times {d_2}[/tex] , where [tex]{d_1},{d_2}[/tex] are two diagonals of rhombus.
Here is given each side of rhombus is 5 cm.
Then the data is inadequate to find.
12. A man bought 5 shirt at Rs. 450 each, 4 trousers at Rs. 750 each and 12 pairs of shoes at Rs. 750 each. The average expenditure per article is
(1) Rs 678.50 (2) Rs 800 (3) Rs 900 (4) Rs 1000
Ans: [tex]\begin{array}{l}
\frac{{(5 \times 450) + (4 \times 750) + (12 \times 750)}}{{5 + 4 + 12}}\\
= \frac{{2250 + 3000 + 9000}}{{21}}\\
= \frac{{14250}}{{21}}\\
= 678.57
\end{array}[/tex]
13. 45% of 280 + 28% of 450 = ?
(1) 152 (2) 252 (3) 354 (4) 454
Ans. [tex]\begin{array}{l}
\frac{{45}}{{100}} \times 280 + \frac{{28}}{{100}} \times 450\\
= \frac{{12600}}{{100}} + \frac{{12600}}{{100}}\\
= 126 + 126\\
= 252
\end{array}[/tex]
14. The radius of a circle is increased by 1%. Then percentage increased in area is
(1) 1% (2) 1.01% (3) 2% (4) 2.01%
Ans. Let the radius of the circle is r unit. Then area of it is [tex]\pi {r^2}uni{t^2}[/tex].
The radius of a circle is increased by 1%. Then new radius is [tex]\frac{{101r}}{{100}}[/tex] unit and area of it is [tex]\pi {(\frac{{101r}}{{100}})^2} = \pi \frac{{10201}}{{10000}}{r^2}uni{t^2}[/tex]
The increased in area is [tex]\pi \frac{{10201}}{{10000}}{r^2} - \pi {r^2} = \pi (\frac{{10201 - 10000}}{{10000}}){r^2} = \pi \frac{{201}}{{10000}}{r^2}uni{t^2}[/tex]
Then percentage increased in area is [tex]\frac{{201 \times 100}}{{10000}} = 2.01[/tex]
15. A dishonest dealer claim to sell his good at the cost price but use a false weight 900 gm for 1 kg. His gain percentage is
(1) 13% (2) [tex]11\frac{1}{9}\% [/tex] (3) 11.25% (4) [tex]12\frac{1}{9}\% [/tex]
Ans. The dishonest dealer weight gain in 900 gm is ( 1000 - 900 ) gm = 100 gm.
The gain percentage in 900 gm is [tex]\frac{{100}}{{900}} \times 100 = \frac{{100}}{9} = 11\frac{1}{9}[/tex]
His gain percentage is [tex]11\frac{1}{9}\% [/tex]
16. By selling a table for Rs. 350 instead of Rs. 400 loss percent increased by 5%; The cost price of table is
(1) Rs. 435 (2) Rs. 417.50 (3) Rs. 1,000 (4) Rs. 1,050
Ans. Let the cost price of the table be Rs. x .
We know the loss %
[tex]\begin{array}{l}
\left( {\frac{{C.P. - S.P.}}{{C.P.}}} \right) \times 100\% \\
\Rightarrow \left( {\frac{{x - 350}}{x}} \right) - \left( {\frac{{x - 400}}{x}} \right) = \frac{5}{{100}}\\
\Rightarrow \frac{{x - 350 - x + 400}}{x} = \frac{5}{{100}}\\
\Rightarrow 5x = 5000\\
\Rightarrow x = 1000
\end{array}[/tex]
Then the cost of the table is Rs.1000
17. A constable is 114 m behind a thief. The constable runs 21 m and thief 15 m in a minute. In what time will the constable catch the thief?
(1) 16 minutes (2) 17 minutes (3) 18 minutes (4) 19 minutes
Ans. The constable covers 21 m in 1 minutes and thief covers 15 m in 1 minutes. Then constable covers ( 21 - 15 ) m = 6 m more than thief in 1 minutes.
Now constable covers 114 m in [tex]\frac{{114}}{6} = 19[/tex] minutes.
In 19 minutes will the constable catch the thief.
18. [tex]{\left[ {{{\left\{ {{{\left( { - \frac{1}{2}} \right)}^2}} \right\}}^{ - 2}}} \right]^{ - 1}}[/tex]
(1) 16 (2) [tex]\frac{1}{{16}}[/tex] (3) 4 (4) [tex]\frac{1}{4}[/tex]
Ans.
[tex]\begin{array}{l}
{\left[ {{{\left\{ {{{\left( { - \frac{1}{2}} \right)}^2}} \right\}}^{ - 2}}} \right]^{ - 1}}\\
= {\left[ {{{\left\{ {\frac{1}{4}} \right\}}^{ - 2}}} \right]^{ - 1}}\\
= {\left[ {16} \right]^{ - 1}}\\
= \frac{1}{{16}}(ans)
\end{array}[/tex]
19. The number of the digits of the square root of 0.00059049 is
(1) 5 (2) 6 (3) 4 (4) 3
Ans. The square root of 0.00059049 is
[tex]\begin{array}{l}
\sqrt {0.00059049} \\
= \sqrt {\frac{{59049}}{{100000000}}} \\
= \frac{{\sqrt {59049} }}{{10000}}\\
= \frac{{243}}{{10000}}\\
= 0.0243
\end{array}[/tex]
Total number of digits is 4
20. 3 men or 5 women can do a work in 12 days. How long will 6 men and 5 women take to finish the work?
(1) 4 days (2) 10 days (3) 15 days (4) 20 days
Ans. From the question we can write 3 men = 5 women. So 6 men and 5 women are total ( 6 + 3 ) men or 9 men.
Then 3 men can do a work in 12 days
Now 1 men can do a work in [tex]12 \times 3 = 36[/tex] days
Therefor 9 men can do this work in [tex]36 \div 9 = 4[/tex] days
21. Two pipes A and B can fill a cistern in 6 minutes and 7 minutes respectively. Both the pipes are opened alternately for 1 minute each what time will they fill the cistern ?
(1) 5 minutes (2) [tex]5\frac{2}{3}[/tex] minutes (3) [tex]6\frac{3}{7}[/tex] minutes (4) [tex]1\frac{1}{4}[/tex] minutes
Ans. Let the volume of the cistern be x.
Then A pipes fill the water in 1 minutes is [tex]\frac{x}{6}[/tex] volume and B pipes fill the water in 1 minutes is [tex]\frac{x}{7}[/tex] volume of cistern.
The volume of cistern fill the water in 2 minutes is [tex]\left( {\frac{x}{6} + \frac{x}{7}} \right) = \left( {\frac{{7 + 6}}{{42}}} \right)x = \frac{{13}}{{42}}x[/tex]
Therefor [tex]\frac{{13}}{{42}}x[/tex] volume of cistern fill of water in 2 minutes.
The part of cistern fill A and B in [tex]2 \times 3 = 6[/tex] minutes = [tex]3 \times \frac{{13x}}{{42}} = \frac{{39x}}{{42}}[/tex] volume.
Part of cistern left =[tex]\left( {x - \frac{{39x}}{{42}}} \right)volume = \frac{{3x}}{{42}}volume = \frac{x}{{14}}volume[/tex]
Now it turn of A
[tex]\frac{x}{6}[/tex] part fill by A in 1 minute.
1 part fill by A in [tex]\frac{6}{x}[/tex] minute.
Then [tex]\frac{x}{{14}}[/tex] part fill by A in [tex]\frac{6}{x} \times \frac{x}{{14}} = \frac{6}{{14}}[/tex] minutes
Total time taken
[tex]\left( {6 + \frac{6}{{14}}} \right) = 6 \times \frac{{15}}{{14}} = \frac{{45}}{7} = 6\frac{3}{7}[/tex]
22. How many times in a day, the hands of a clock are straight ?
(1) 22 (2) 24 (3) 44 (4) 48
Ans. Hands of a clock point in opposite directions i,e in the same straight lines in 11 times in every 12 hours. Hence in a day hands of clock are 22 times in straight lines.
23. The reflex angle between the hands of a clock at 10.25 is
(1) [tex]{180^0}[/tex] (2) [tex]192\frac{{{1^0}}}{2}[/tex] (3) [tex]{195^0}[/tex] (4) [tex]197\frac{{{1^0}}}{2}[/tex]
Ans. Angle subtended by hour hand in [tex]10\frac{{25}}{{60}} = \frac{{125}}{{12}}[/tex] hrs = [tex]\left( {\frac{{{{360}^0}}}{{12}} \times \frac{{125}}{{12}}} \right) = 312\frac{{{1^0}}}{2}[/tex]
Angle subtended by minute hand in 25 min = [tex]\left( {\frac{{{{360}^0}}}{{60}} \times 25} \right) = {150^0}[/tex]
Reflex angle
[tex]\begin{array}{l}
{360^0} - {\left( {312.5 - 150} \right)^0}\\
= {360^0} - {162.5^0}\\
= {197.5^0}\\
= 197\frac{{{1^0}}}{2}
\end{array}[/tex]
24. The sum 5 + 6 + 7 + 8 + ............ + 19 = ?
(1) 150 (2) 170 (3) 180 (4) 190
Ans. We know the sum of A.P is
[tex]\frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right][/tex]
Where a = 5, n = 15, d = 1
Sum [tex]\frac{{15}}{2}\left[ {2 \times 5 + \left( {15 - 1} \right) \times 1} \right] = \frac{{15}}{2}\left( {10 + 14} \right) = \frac{{15 \times 24}}{2} = 180[/tex]
25. [tex]0.\dot 3{\rm{ }} + {\rm{ }}0.\dot 4{\rm{ }} + {\rm{ }}0.\dot 7{\rm{ }} + {\rm{ }}0.\dot 8{\rm{ }} = {\rm{ }}?[/tex]
(1) 2.4444 (2) 2.44 (3) 2.444 (4) [tex]2.\dot 4{\rm{ }}[/tex]
Ans.
[tex]0.\dot 3{\rm{ }} + {\rm{ }}0.\dot 4{\rm{ }} + {\rm{ }}0.\dot 7{\rm{ }} + {\rm{ }}0.\dot 8{\rm{ }} [/tex]
[tex]= \frac {3}{9} + \frac {4}{9} + \frac {7}{9} + \frac {8}{9} [/tex]
[tex]= \frac {{22}}{9}[/tex]
[tex]= 2.444[/tex]
[tex]= 2.\dot 4{\rm{ }}[/tex]
26. If the ratio of three numbers is 3 : 4 : 5 and their LCM is 1200, then the smaller number is
(1) 60 (2) 80 (3) 100 (4) 120
Ans. Let the three numbers are 3x , 4x and 5x.
Therefor the LCM of them is [tex]3 \times 4 \times 5 \times x = 60x[/tex]
Then we can say
[tex]\begin{array}{l} 60x = 1200\\ \Rightarrow x = 20 \end{array}[/tex]
Then smaller number is [tex]3 \times 20 = 60[/tex]
27. The product of two numbers is 1575 and its division is [tex]\frac{9}{7}[/tex]. Then the two numbers are
(1) 45 , 35 (2) 81 , 63 (3) 35 , 27 (4) 36 , 35
Ans. Let two numbers are x and y.
Then [tex]x \times y = 1575.........................(i)[/tex]
and [tex]\frac{x}{y} = \frac{9}{7}.....................................(ii)[/tex]
From (ii) we get [tex]\frac{x}{y} = \frac{9}{7} \Rightarrow x = \frac{{9y}}{7}................................\left( {iii} \right)[/tex]
put (iii) in (i)
[tex]\begin{array}{l}
\frac{{9y}}{7} \times y = 1575\\
\Rightarrow {y^2} = \frac{{1575 \times 7}}{9} = 1225\\
\Rightarrow y = \sqrt {1225} = \pm 35
\end{array}[/tex]
Then another number will be [tex]1575 \div \left( { \pm 35} \right) = \pm 45[/tex]
28. [tex]\sqrt {2025} + \sqrt {441} + \sqrt {169} = ?[/tex]
(1) 59 (2) 69 (3) 79 (4) 89
Ans.
[tex]\begin{array}{l}
\sqrt {2025} + \sqrt {441} + \sqrt {169} \\
= \sqrt {5 \times 5 \times 9 \times 9} + \sqrt {9 \times 7 \times 7} + \sqrt {13 \times 13} \\
= 5 \times 9 + 3 \times 7 + 13\\
= 45 + 21 + 13\\
= 79
\end{array}[/tex]
29. The ratio of two numbers is 5 : 8 and their difference is 69. Then the two numbers are
(1) 69 , 128 (2) 115 , 184 (3) 43 , 112 (4) 128 , 197
Ans. Let the two numbers are 5x and 8x. Then we can write
8x - 5x = 69
or, 3x = 69
or, x = 23
Therefor two numbers are [tex]\left( {5 \times 23} \right) = 115[/tex] and [tex]\left( {8 \times 23} \right) = 184[/tex]
30. If [tex]A:B = \frac{1}{6}:\frac{1}{5},B:C = \frac{1}{4}:\frac{1}{3},C:D = \frac{1}{3}:\frac{1}{5}[/tex] then A : D is
(1) [tex]\frac{1}{{24}}:\frac{1}{{25}}[/tex] (2) [tex]\frac{1}{{27}}:\frac{1}{{25}}[/tex] (3) [tex]\frac{1}{{24}}:\frac{1}{{29}}[/tex] (4) None of the above
Ans.
[tex]\frac{A}{B} = \frac{5}{6},\frac{B}{C} = \frac{3}{4},\frac{C}{D} = \frac{5}{3}[/tex]
Then
[tex]\begin{array}{l} \frac{A}{B} \times \frac{B}{C} \times \frac{C}{D} = \frac{5}{6} \times \frac{3}{4} \times \frac{5}{3}\\
\Rightarrow \frac{A}{D} = \frac{{25}}{{24}}\\ \Rightarrow A:D = \frac{1}{{24}}:\frac{1}{{25}} \end{array}[/tex]
31. [tex]\frac{{{{\left( {6.5} \right)}^2} - {{\left( {3.15} \right)}^2}}}{{\left( {6.5 + 3.15} \right)}} = ?[/tex]
(1) 3.5 (2) 3.51 (3) 3.52 (4) 3.35
Ans.
[tex]\begin{array}{l}
\frac{{{{\left( {6.5} \right)}^2} - {{\left( {3.15} \right)}^2}}}{{\left( {6.5 + 3.15} \right)}}\\
= \frac{{\left( {6.5 + 3.15} \right) \times \left( {6.5 - 3.15} \right)}}{{\left( {6.5 + 3.15} \right)}}\\
= \left( {6.5 - 3.15} \right)\\
= 3.35
\end{array}[/tex]u
32. The compound interest on Rs. 2000 for 2 years at 8%per annum is
(1) Rs. 220.80 (2) Rs. 232.80 (3) Rs. 332.80 (4) Rs. 532.80
Ans. We know that [tex]C.I. = P\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^n} - 1} \right][/tex]
Where P = 2000 , r = 8 , n = 2
[tex]\begin{array}{l}
C.I. = 2000\left[ {{{\left( {1 + \frac{8}{{100}}} \right)}^2} - 1} \right]\\
= 2000\left[ {{{\left( {\frac{{108}}{{100}}} \right)}^2} - 1} \right]\\
= 2000\left( {\frac{{108}}{{100}} + 1} \right)\left( {\frac{{108}}{{100}} - 1} \right)\\
= 2000 \times \frac{{208}}{{100}} \times \frac{8}{{100}}\\
= 332.80
\end{array}[/tex]
33. The rational numbers lying between [tex]\frac{1}{4}[/tex] and [tex]\frac{3}{4}[/tex] are
(1) [tex]\frac{9}{{40}},\frac{{31}}{{41}}[/tex] (2) [tex]\frac{{13}}{{50}},\frac{{264}}{{350}}[/tex] (3) [tex]\frac{{63}}{{250}},\frac{{187}}{{250}}[/tex] (4) [tex]\frac{{262}}{{1000}},\frac{{752}}{{1000}}[/tex]
Ans. [tex]\frac{1}{4} \times \frac{{250}}{{250}} = \frac{{250}}{{1000}},\frac{3}{4} \times \frac{{250}}{{250}} = \frac{{750}}{{1000}}[/tex]
Now from the option
[tex]\begin{array}{l}
\frac{{63}}{{250}} \times \frac{4}{4} = \frac{{252}}{{1000}},\frac{{187}}{{250}} \times \frac{4}{4} = \frac{{748}}{{1000}}\\
i.e\frac{1}{4} < \frac{{63}}{{250}} < \frac{3}{4},\frac{1}{4} < \frac{{187}}{{250}} < \frac{3}{4}
\end{array}[/tex]
Then option (3) lying between [tex]\frac{1}{4}[/tex] and [tex]\frac{3}{4}[/tex]
34. [tex]{\left( {77 + \frac{1}{{77}}} \right)^2} - {\left( {77 - \frac{1}{{77}}} \right)^2} = ?[/tex]
(1) 2 (2) 4 (3) 1 (4) 77
Ans.
[tex]\begin{array}{l}
{\left( {77 + \frac{1}{{77}}} \right)^2} - {\left( {77 - \frac{1}{{77}}} \right)^2}\\
= 4 \times 77 \times \frac{1}{{77}}\left[ {4ab = {{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}} \right]\\
= 4
\end{array}[/tex]
35. By what least number must 21600 be multiplied to make it a perfect cube?
(1) 6 (2) 10 (3) 30 (4) 60
Ans. [tex]21600 = 3 \times 3 \times 3 \times 2 \times 2 \times 2 \times 10 \times 10[/tex]
At least 10 multiplied to this number to make a perfect cube.
36. If P = 50% of Q and Q = 50% of R, then P : Q : R = ?
(1) 1 : 2 : 4 (2) 1 : 4 : 2 (3) 4 : 2 : 1 (4) 2 : 1 : 4
Ans. P = 50% of Q, then [tex]P = \frac{{50}}{{100}}Q = \frac{1}{2}Q[/tex] ...........................(i)
and Q = 50% of R, then [tex]Q = \frac{{50}}{{100}}R = \frac{1}{2}R[/tex] .............................(ii)
From (i) 2P=Q , and (ii) 2Q=R
Then [tex]2 \times 2P = R \Rightarrow 4P = R[/tex]
Therefor P : Q : R = P : 2P : 4P =1 : 2 ; 4
37. The average of first 100 natural numbers is
(1) 50 (2) 50.5 (3) 51 (4) 51.5
Ans. We know that the average of first n natural numbers is [tex]\frac{{\left( {n + 1} \right)}}{2}[/tex]
Then the required average is [tex]\frac{{\left( {100 + 1} \right)}}{2} = \frac{{101}}{2} = 50.5[/tex]
38. If the sum of two numbers is 10 and the sum their reciprocals is [tex]\frac{5}{{12}}[/tex] the numbers would be
(1) ( 8 , 2 ) (2) ( 6 , 4 ) (3) ( 7 , 3 ) (4) ( 9 , 1 )
Ans. Let the two numbers are x and y .
Then x + y = 10 ......................(i) and [tex]\frac{1}{x} + \frac{1}{y} = \frac{5}{{12}}..........................(ii)[/tex]
Now from (i)
x = 10 - y...............................(iii)
use (iii) in (ii) we get
[tex]\begin{array}{l}
\frac{1}{{10 - y}} + \frac{1}{y} = \frac{5}{{12}}\\
\Rightarrow \frac{{y + 10 - y}}{{y\left( {10 - y} \right)}} = \frac{5}{{12}}\\
\Rightarrow \frac{{10}}{{y\left( {10 - y} \right)}} = \frac{5}{{12}}\\
\Rightarrow 24 = 10y - {y^2}\\
\Rightarrow {y^2} - 10y + 24 = 0\\
\Rightarrow {y^2} - \left( {6 + 4} \right)y + 24 = 0\\
\Rightarrow y\left( {y - 6} \right) - 4\left( {y - 6} \right) = 0\\
\Rightarrow \left( {y - 6} \right)\left( {y - 4} \right) = 0\\
\Rightarrow y = 6,4
\end{array}[/tex]
If y = 6 , then x = 4 and if y = 4 , then x = 6
39. The product of two successive numbers is 1980. The smaller number is
(1) 34 (2) 35 (3) 44 (4) 45
Ans. Let n and (n+1) are two successive numbers. Then
[tex]\begin{array}{l}
n\left( {n + 1} \right) = 1980\\
\Rightarrow {n^2} + n - 1980 = 0\\
\Rightarrow {n^2} + \left( {45 - 44} \right)n - 1980 = 0\\
\Rightarrow {n^2} + 45n - 44n - 1980 = 0\\
\Rightarrow n\left( {n + 45} \right) - 44\left( {n + 45} \right) = 0\\
\Rightarrow \left( {n - 44} \right)\left( {n + 45} \right) = 0\\
\Rightarrow n = 44,\left( { - 45} \right)
\end{array}[/tex]
Then smaller number is 44
40. [tex]\frac{4}{5}[/tex] of certain number is 64. Half of the number
(1) 32 (2) 40 (3) 80 (4) 16
Ans. Let the certain number is x. Then
[tex]\begin{array}{l}
\frac{4}{5}x = 64\\
\Rightarrow x = 64 \times \frac{5}{4} = 80
\end{array}[/tex]
Therefor the half of x is [tex]80 \div 2 = 40[/tex]
41. Which is greater [tex]\sqrt 2 [/tex] or [tex]3\sqrt 3 [/tex] ?
(1) [tex]\sqrt 2 [/tex] (2) [tex]3\sqrt 3 [/tex] (3) Two are equal (4) None of the above
Ans. Now to prove this we subtract them. Now
[tex]\begin{array}{l}
\sqrt 2 - 3\sqrt 3 \\
= \frac{{\left( {\sqrt 2 - 3\sqrt 3 } \right)\left( {\sqrt 2 + 3\sqrt 3 } \right)}}{{\left( {\sqrt 2 + 3\sqrt 3 } \right)}}\\
= \frac{{2 - 9 \times 3}}{{\left( {\sqrt 2 + 3\sqrt 3 } \right)}}\\
= \frac{{2 - 27}}{{\left( {\sqrt 2 + 3\sqrt 3 } \right)}} < 0\left[ {as 27 > 2} \right]
\end{array}[/tex]
The greater number is [tex]3\sqrt 3 [/tex]
42. The fraction equivalent to [tex]\frac{2}{5}\% [/tex] is
(1) [tex]\frac{1}{{40}}[/tex] (2) [tex]\frac{1}{{125}}[/tex] (3) [tex]\frac{1}{{250}}[/tex] (4) [tex]\frac{1}{{500}}[/tex]
Ans. [tex]\frac{2}{5}\% = \frac{2}{5} \times \frac{1}{{100}} = \frac{1}{{250}}[/tex]
43. If, [tex]\frac{1}{5}:\frac{1}{x}::\frac{1}{x}:\frac{1}{{1.25}}[/tex] , then x = ?
(1) 1.5 (2) 2 (3) 2.5 (4) 3.5
Ans.
[tex]\begin{array}{l}
\frac{1}{5}:\frac{1}{x}::\frac{1}{x}:\frac{1}{{1.25}}\\
\Rightarrow \frac{{{\textstyle{1 \over 5}}}}{{{\textstyle{1 \over x}}}} = \frac{{{\textstyle{1 \over x}}}}{{{\textstyle{1 \over {1.25}}}}}\\
\Rightarrow {\left( {\frac{1}{x}} \right)^2} = \frac{1}{5} \times \frac{1}{{1.25}}\\
\Rightarrow {\left( {\frac{1}{x}} \right)^2} = \frac{1}{{6.25}}\\
\Rightarrow \frac{1}{x} = \sqrt {\frac{{100}}{{625}}} = \sqrt {\frac{{10 \times 10}}{{25 \times 25}}} = \frac{{10}}{{25}}\\
\Rightarrow x = \frac{{25}}{{10}} = 2.5
\end{array}[/tex]
44. By selling 100 pencils, a shopkeeper gains the selling price of 20 pencils. His gain percentage is
(1) 25% (2) 20% (3) 15% (4) 12%
Ans. Let selling price of 1 pencil is Rs. k .
Therefore the selling price of 100 pencils is Rs. 100k.
From the question the profit = selling price of 20 pencils = Rs. 20k.
Therefor the cost of 100 pencils is ( 100k - 20k ) = 80 k.
We know
[tex]\begin{array}{l}
Gain\% = \frac{{S.P - C.P}}{{C.P}} \times 100\\
= \frac{{100k - 80k}}{{80k}} \times 100\\
= \frac{{20k}}{{80k}} \times 100\\
= 25\%
\end{array}[/tex]
45. Two numbers are in the ratio of 3 : 5. If each number is increased by 10, the ratio becomes 5 : 7. The numbers are
(1) 3 , 5 (2) 7 , 9 (3) 13 , 22 (4) 15 , 25
Ans. Let two numbers are x and y.
From the question we get
[tex]\begin{array}{l}
x:y = 3:5 \Rightarrow \frac{x}{y} = \frac{3}{5}.........................(i)\\
\left( {x + 10} \right):\left( {y + 10} \right) = 5:7 \Rightarrow \frac{{\left( {x + 10} \right)}}{{\left( {y + 10} \right)}} = \frac{5}{7}.....................(ii)
\end{array}[/tex]
From (i) we get [tex]x = \frac{3}{5}y[/tex]................................(iii)
use (iii) in (i)
[tex]\begin{array}{l}
\frac{{\left( {x + 10} \right)}}{{\left( {y + 10} \right)}} = \frac{5}{7}\\
\Rightarrow \frac{{\left( {\frac{3}{5}y + 10} \right)}}{{\left( {y + 10} \right)}} = \frac{5}{7}\\
\Rightarrow 7\left( {\frac{3}{5}y + 10} \right) = 5\left( {y + 10} \right)\\
\Rightarrow 21y + 350 = 25y + 250\\
\Rightarrow 4y = 100\\
\Rightarrow y = 25
\end{array}[/tex]
Therefor [tex]x = \frac{3}{5} \times 25 = 15[/tex]
46. The simplified values of
[tex]\frac{1}{{56}} + \frac{1}{{72}} + \frac{1}{{90}} + \frac{1}{{110}} + \frac{1}{{132}}[/tex]
(1) [tex]\frac{1}{{84}}[/tex] (2) [tex]\frac{1}{{28}}[/tex] (3) [tex]\frac{5}{{84}}[/tex] (4) [tex]\frac{1}{{12}}[/tex]
Ans.
[tex]\begin{array}{l}
\frac{1}{{56}} + \frac{1}{{72}} + \frac{1}{{90}} + \frac{1}{{110}} + \frac{1}{{132}}\\
= \frac{1}{{7 \times 8}} + \frac{1}{{8 \times 9}} + \frac{1}{{9 \times 10}} + \frac{1}{{10 \times 11}} + \frac{1}{{11 \times 12}}\\
= \left( {\frac{1}{7} - \frac{1}{8}} \right) + \left( {\frac{1}{8} - \frac{1}{9}} \right) + \left( {\frac{1}{9} - \frac{1}{{10}}} \right) + \left( {\frac{1}{{10}} - \frac{1}{{11}}} \right) + \left( {\frac{1}{{11}} - \frac{1}{{12}}} \right)\\
= \frac{1}{7} - \frac{1}{8} + \frac{1}{8} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{11}} + \frac{1}{{11}} - \frac{1}{{12}}\\
= \frac{1}{7} - \frac{1}{{12}}\\
= \frac{{12 - 7}}{{84}} = \frac{5}{{84}}
\end{array}[/tex]
47. [tex]7 \times 0.7 \times 0.07 \times 7000 = ?[/tex]
(1) 24.01 (2) 2.401 (3) 240.1 (4) 2401
Ans.
[tex]\begin{array}{l}
7 \times 0.7 \times 0.07 \times 7000\\
= 7 \times \frac{7}{{10}} \times \frac{7}{{100}} \times 7000\\
= 2401
\end{array}[/tex]
48. The value of [tex]\frac{3}{4}[/tex] property is Rs. 21,000, the value of [tex]\frac{4}{7}[/tex] of the same property is
(1) Rs. 16,000 (2) Rs. 28,000 (3) Rs. 14,000 (4) Rs. 19,000
Ans. Let the value of property be x.
Then
[tex]\begin{array}{l}
\frac{3}{4}x = 21000\\
\Rightarrow x = 21000 \times \frac{4}{3}\\
\Rightarrow x = 28000
\end{array}[/tex]
Therefor the value of [tex]\frac{4}{7}[/tex] of the same property is [tex]\frac{4}{7} \times 28000 = 16000[/tex]
49. The bell of a wall-clock requires 3 seconds to ring 5 times. The time required for the bell to ring 6 times is
(1) 4 seconds (2) [tex]3\frac{3}{4}[/tex] (3) 5 seconds (4) [tex]3\frac{1}{2}[/tex]
Ans. There are 4 gaps between 5 times ring. The gaps are shown that ._._._._.
Therefor 4 gaps take 3 seconds
Then 1 gaps take [tex]\frac{3}{4}[/tex]
The number of gaps 6 times ring are ._._._._._. = 5
The time required for the bell to ring 6 times is [tex]\frac{3}{4} \times 5 = \frac{{15}}{4} = 3\frac{3}{4}[/tex] seconds
50. [tex]1.\dot 3\dot 5 \div 2.\dot 0\dot 3 = ?[/tex]
(1) [tex]0.\dot 3[/tex] (2) [tex]0.\dot 5[/tex] (3) [tex]0.\dot 6[/tex] (4) [tex]0.\dot 7[/tex]
Ans. [tex]1.\dot 3\dot 5 \div 2.\dot 0\dot 3[/tex]
[tex]1.\dot 3\dot 5 = \frac{{135 - 1}}{{990}} = \frac{{134}}{{990}}[/tex]
[tex]2.\dot 0\dot 3 = \frac{{203 - 2}}{{990}} = \frac{{201}}{{990}}[/tex]
[tex]1.\dot 3\dot 5 \div 2.\dot 0\dot 3 = \frac{{134}}{{201}} = 0.666666666666666 = 0.\dot 6[/tex]
51. In what proportion must water be added to spirit to gain 20% by selling it at the cost price?
(1) 2 : 5 (2) 1 : 5 (3) 3 : 5 (4) 4 : 5
Ans. Let the quantity of spirit be x liter and the quantity of water be y liter.
Let per liter C.P. of spirit = Rs. 1
i.e C.P. of x liter = Rs. x.
And S.P. of (x+1) liter = Rs. ( 1 + x )
[ since the mixture is sold on C.P. rate ]
Now gain is 20%. So from S.P. = C.P.[tex]\left( {1 + \frac{{gain\% }}{{100}}} \right)[/tex]
[tex]\begin{array}{l}
x + y = x\left( {1 + \frac{{20}}{{100}}} \right)\\
\Rightarrow y = x\left( {1 + \frac{1}{5}} \right) - x\\
\Rightarrow y = x\left( {1 + \frac{1}{5} - 1} \right)\\
\Rightarrow y = x\frac{1}{5}\\
\Rightarrow \frac{y}{x} = \frac{1}{5}\\
\Rightarrow y:x = 1:5
\end{array}[/tex]
i.e water spirit = 1 : 5
52. [tex]{\left( { 0.\bar 1} \right)^2}\left[ {1 - 9 \times {{\left( {0. 1\bar 6} \right)}^2}} \right] = ?[/tex]
(1) [tex]\frac{1}{{162}}[/tex] (2) [tex]\frac{1}{{108}}[/tex] (3) [tex]\frac{1}{{109}}[/tex] (4) [tex]\frac{{7696}}{{{{10}^6}}}[/tex]
Ans.
\[\begin{array}{l}
{\left( {0.\bar 1} \right)^2}\left[ {1 - 9 \times {{\left( {0.1\bar 6} \right)}^2}} \right]\\
= {\left( {\frac{1}{9}} \right)^2}\left[ {1 - 9 \times {{\left( {\frac{{16 - 1}}{{90}}} \right)}^2}} \right]\\
= \frac{1}{{81}}\left[ {1 - 9 \times {{\left( {\frac{{15}}{{90}}} \right)}^2}} \right]\\
= \frac{1}{{81}}\left[ {1 - 9 \times \frac{{225}}{{8100}}} \right]\\
= \frac{1}{{81}}\left[ {1 - \frac{1}{4}} \right]\\
= \frac{1}{{81}} \times \frac{3}{4} = \frac{1}{{27 \times 4}}\\
= \frac{1}{{108}}
\end{array}\]
53. Find the odd one out:
(1) 27 (2) 64 (3) 81 (4) 125
Ans. Now [tex]27 = {3^3},64 = {4^3},81 = {3^4},125 = {5^3}[/tex]
We see option (1) 27, option (2) 64 and option (4) 125 are cube of 3 , 4 and 5 respectively and option (3) 81 is not of them .Therefore option (3) 81 is odd.
54. Rs. 6,400 are divided among three workers in the ratio [tex]\frac{3}{5}:2:\frac{5}{3}[/tex] . The share of the second worker is
(1) Rs. 2,560 (2) Rs. 3,000 (3) Rs. 3,200 (4) Rs. 3,840
Ans. Let the workers pay the money are [tex]\frac{3}{5}x,2x,\frac{5}{3}x[/tex] respectively. Then from the question we get
[tex]\begin{array}{l}
\frac{3}{5}x + 2x + \frac{5}{3}x = 6400\\
\Rightarrow \frac{{9x + 30x + 25x}}{{15}} = 6400\\
\Rightarrow \frac{{64x}}{{15}} = 6400\\
\Rightarrow x = 6400 \times \frac{{15}}{{64}} = 1500
\end{array}[/tex]
The share of the second worker is \[2x = 2 \times 1500 = 3000\] rupees
55. A, B, C started a shop by investing Rs. 27,000, Rs. 81,000 and Rs. 72,000 respectively. At the end of the year, B's share of total profit was Rs. 36,000. The total profit was
(1) Rs. 1,08,000 (2) Rs. 1,16,000 (3) Rs. 80,000 (4) Rs. 92,000
Ans. Let total profit is Rs. x.
B's share of total profit = [tex]\left( {\frac{{81000}}{{27000 + 81000 + 72000}}} \right) \times x[/tex]
[tex]\begin{array}{l}
\Rightarrow 36000 = \left( {\frac{{81000}}{{27000 + 81000 + 72000}}} \right) \times x\\
\Rightarrow 36000 = \left( {\frac{{81000}}{{180000}}} \right)x\\
\Rightarrow x = 36000 \times \frac{{180000}}{{81000}} = 4 \times 20000 = 80000
\end{array}[/tex]
The total profit is Rs. 80,000.
56. If [tex]\sqrt {15} = 3.8729[/tex] , then the value of [tex]\frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}[/tex] is
(1) 7.8729 (2) 7.7829 (3) 7.2987 (4) 7.8279
Ans.
[tex]\begin{array}{l}
\frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}\\
= \frac{{\left( {\sqrt 5 + \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 - \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}}\\
= \frac{{{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{{5 - 3}}\\
= \frac{{5 + 3 + 2\sqrt {5 \times 3} }}{2}\\
= \frac{{8 + 2\sqrt {15} }}{2}\\
= 4 + \sqrt {15} \\
= 4 + 3.8729\\
= 7.8729
\end{array}[/tex]
57. Find the odd one out :
(1) Ear (2) Eye (3) Tongue (4) Blood
Ans. Ear , Eye and tongue all are our sense organs but Blood is not . Then Blood is odd.
58. A man performs [tex]\frac{2}{{15}}[/tex] of the total journey by rail , [tex]\frac{9}{{20}}[/tex] by bus and remaining 10 km on cycle. His total journey is
(1) 30.2 km (2) 38.4 km (3) 23 km (4) 24 km
Ans. Let the total journey is x km.
Then his rail journey is [tex]\frac{2}{{15}}x[/tex] km. His bus journey is [tex]\frac{9}{{20}}x[/tex] km.
Then
[tex]\begin{array}{l}
\frac{2}{{15}}x + \frac{9}{{20}}x + 10 = x\\
\Rightarrow \frac{{8x + 27x - 60x}}{{60}} = - 10\\
\Rightarrow \frac{{35x - 60x}}{{60}} = - 10\\
\Rightarrow - \frac{{25x}}{{60}} = - 10\\
\Rightarrow x = \frac{{10 \times 60}}{{25}} = 24
\end{array}[/tex]
Total journey is 24 km.
59. 40 men take 8 days to earn Rs. 2000. How many men will earn Rs. 200 in 2 days?
(1) 10 (2) 12 (3) 14 (4) 1
Ans. Wage of one man in a day in 1 day = [tex]\frac{{2000}}{{40 \times 8}} = Rs.\frac{{50}}{8} = Rs.\frac{{25}}{4}[/tex]
Wage of x men in a day = [tex]\frac{{25}}{4}x[/tex]
Wage of x men in 2 days
[tex]\begin{array}{l}
\frac{{25}}{4} \times 2x = \frac{{25}}{2}x\\
\Rightarrow \frac{{25}}{2}x = 200\\
\Rightarrow x = \frac{{200 \times 2}}{{25}}\\
\Rightarrow x = 16
\end{array}[/tex]
16 men will earn Rs. 200 2 days.
60. The least perfect square number which is divisible by 3 , 4 , 5 , 6 and 8 is
(1) 900 (2) 1200 (3) 1600 (4) 3600
Ans. The L.C.M of 3 , 4 , 5 , 6 and 8 is [tex]3 \times 4 \times 5 \times 2 = 120[/tex] out of these options only 3600 is the perfect square number which is divisible by 120.
61. Fill in the blanks in the sequence 3 , 15 , 4 , 16 , 5 , 17 , 6 , _ , 7.
(1) 12 (2) 18 (3) 15 (4) 13
Ans. Since the difference between 1st and 2nd is 12 , 3rd and 4th is 12 , 5th and 6th is 12. Then the difference between 7th and 8th should be always 12. Therefore 8th number is 6 + 12 = 18.
62. The greatest possible length which can be use to measure exactly the length 7 m , 3 m 85 cm , 12 m 95 cm is
(1) 15 cm (2) 25 cm (3) 35 cm (4) 42 cm
Ans. Required length = H.C.F of 7 m , 3 m 85 cm , 12 m 95 cm i.e H.C.F of 700 cm 385 cm and 1295 cm.
Hence H.C.F is 35 cm . Therefore the required possible length will be 35 cm.
63. [tex]\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right)\left( {1 - \frac{1}{5}} \right)..................\left( {1 - \frac{1}{{99}}} \right)\left( {1 - \frac{1}{{100}}} \right) = ?[/tex]
(1) [tex]\frac{1}{{100}}[/tex] (2) [tex]\frac{{99}}{{100}}[/tex] (3) [tex]\frac{1}{{50}}[/tex] (4) [tex]\frac{1}{{25}}[/tex]
Ans.
[tex]\begin{array}{l}
\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right)\left( {1 - \frac{1}{5}} \right).........................\left( {1 - \frac{1}{{99}}} \right)\left( {1 - \frac{1}{{100}}} \right)\\
= \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times .......................\frac{{98}}{{99}} \times \frac{{99}}{{100}}\\
= \frac{2}{{100}}\\
= \frac{1}{{50}}
\end{array}[/tex]
64. Find the odd one out:
(1) Motor-cycle (2) Scooter (3) Bi-cycle (4) Van
Ans. Bicycle is odd one out . Motor-cycle , Scooter and Van run on fuel i.e petrol. But cycle dose not require fuel or petrol.
65. What number should come next in the sequence 19 , 2 , 38 , 3 , 114 , 4 , _ ?
(1) 228 (2) 256 (3) 356 (4) 456
Ans. Since 3rd number = 1st number multiply 2nd number [tex] = 19 \times 2 = 38[/tex]
5th number = 3rd number multiply 4th number [tex] = 38 \times 3 = 114[/tex]
Then 7th number = 5th number multiply 6th number [tex] = 114 \times 4 = 456[/tex]
66. What should come next in the sequence 0 , 2 , 8 , 14 , 24 , 34 , _ ?
(1) 48 (2) 42 (3) 40 (4) 38
Ans. The different between 1st term and 2nd term is 2 .
The different between 2nd term and 3rd term is 6 .
The different between 3rd term and 4th term is 6
The different between 4th term and 5th term is 10 .
The different between 5th term and 6th term is 10
Since the different between 6th and 7th term will be 14. Hence the number will be 34 + 14 = 38.
67. Which of the following fraction is the largest ?
(1) [tex]\frac{7}{8}[/tex] (2) [tex]\frac{{13}}{{16}}[/tex] (3) [tex]\frac{{31}}{{40}}[/tex] (4) [tex]\frac{{63}}{{80}}[/tex]
Ans. Now L.C.M of 8 , 16 , 40 and 80 is 80.
Now [tex]\frac{7}{8} = \frac{{7 \times 10}}{{8 \times 10}} = \frac{{70}}{{80}}[/tex]
[tex]\frac{{13}}{{16}} = \frac{{13 \times 5}}{{16 \times 5}} = \frac{{65}}{{80}}[/tex]
[tex]\frac{{31}}{{40}} = \frac{{31 \times 2}}{{40 \times 2}} = \frac{{62}}{{80}}[/tex]
[tex]\frac{{63}}{{80}} = \frac{{63 \times 1}}{{80 \times 1}} = \frac{{63}}{{80}}[/tex]
We see that option (1) is largest of them.
68. Fill the blank in the series looking at both the letter pattern: [tex]{B_2}CD,\_,BC{D_4},{B_5}CD,B{C_6}D[/tex]
(1) [tex]{B_2}{C_2}D[/tex] (2) [tex]B{C_3}D[/tex] (3) [tex]{B_2}{C_3}D[/tex] (4) [tex]BC{D_7}[/tex]
Ans. In every next step the numeral attached with next letter and increases by 1.
[tex]{B_2}CD,(B{C_3}D),BC{D_4},{B_5}CD,B{C_6}D[/tex]
69. If [tex]\frac{1}{{25}}:\frac{1}{x}::\frac{1}{{{x^2}}}:\frac{1}{{78.125}}[/tex] then x = ?
(1) 1.5 (2) 2 (3) 12.5 (4) 3.5
Ans.
[tex]\begin{array}{l}
\frac{1}{{25}}:\frac{1}{x}::\frac{1}{{{x^2}}}:\frac{1}{{78.125}}\\
\Rightarrow \frac{x}{{25}} = \frac{{78.125}}{{{x^2}}}\\
\Rightarrow {x^3} = 25 \times 78.125\\
\Rightarrow {x^3} = \frac{{25 \times 78125}}{{1000}} = \frac{{25 \times 25 \times 25 \times 125}}{{1000}}\\
\Rightarrow x = \sqrt[3]{{\frac{{25 \times 25 \times 25 \times 125}}{{1000}}}} = \frac{{25 \times 5}}{{10}} = 12.5
\end{array}[/tex]
70. The sum of all exterior angles of the convex polynomial of n sides is
(1) 4 right angle (2) [tex]\frac{2}{n}[/tex] right angle (3) 2 ( n - 2 ) right angle (4) [tex]\frac{2}{n}[/tex] right angle
Ans. Sum of exterior angles of any polynomial is 4 right angles [ i.e [tex]{360^0}[/tex] ]
71. Look carefully for the pattern, and then choose which pair of numbers comes next:
1 ,10 , 7 , 20 , 13 , 30 , 19
(1) 25, 40 (2) 40, 25 (3) 25, 31 (4) 40, 50
Ans. Since the common difference between 1 , 7 , 13 , 19 is 6 and the common difference between 10 , 20 , 30 is 10 .Then then the pair which comes next is ( 30 + 10 , 19 + 6 ) = ( 40 , 25 )
72. What number should come next in the sequence
3 , 8 , 15 , 24 , 35 , _ ?
(1) 44 (2) 46 (3) 48 (4) 50
Ans. The difference between 1st and 2nd term is 5, 2nd and 3rd term is 7, 3rd and 4th term is 9, 4th and 5th term is 11.
Then also the difference between 5th and 6th term will be 13. Then the next number will be ( 35 + 13 ) = 48 .
73. If the diagonals of the rhombus are 8 cm and 6 cm, the square of its side is
(1) [tex]25c{m^2}[/tex] (2) [tex]24c{m^2}[/tex] (3) [tex]55c{m^2}[/tex] (4) [tex]36c{m^2}[/tex]
Ans. We know the diagonals of rhombus bisect each other at right angle.
Then the square of rhombus side is
[tex]\begin{array}{l}
[{\left( {\frac{8}{2}} \right)^2} + {\left( {\frac{6}{2}} \right)^2}]c{m^2}\\
= [{\left( 4 \right)^2} + {\left( 3 \right)^2}]c{m^2}\\
= [16 + 9]c{m^2}\\
= 25c{m^2}
\end{array}[/tex]
74. What comes next in the sequence 1 , 3 , 7 , 15 , 31 , 63 , _ ?
(1) 127 (2) 125 (3) 121 (4) 129
Ans. Since 3 = ( 1 + 2 ) i.e 2nd term = ( 1st term + [tex]{2^1}[/tex] )
7 = ( 3 + 4 ) i.e 3rd term = (2nd term + [tex]{2^2}[/tex] )
15 = ( 7 + 8 ) i.e 4th term = (3rd term + [tex]{2^3}[/tex] )
31 = ( 15 + 16 ) i.e 5th term = ( 4th term + [tex]{2^4}[/tex] )
63 = ( 31 + 32 ) i.e 6th term = ( 5th term + [tex]{2^5}[/tex] )
Therefore 7th term = ( 6th term + [tex]{2^6}[/tex] ) = ( 63 + 64 ) = 127
75. Find the odd one out:
(1) Rocket (2) Aeroplane (3) Helicopter (4) Van-rickshaw
Ans. We see that Rocket, Aeroplane and Helicopter all fly in the sky but Van-rickshaw runs on road. So Van-rickshaw is odd one.
76. The difference of [tex]1\frac{3}{{16}}[/tex] and its reciprocal is
(1) [tex]1\frac{1}{8}[/tex] (2) [tex]1\frac{1}{3}[/tex] (3) [tex]\frac{{15}}{{16}}[/tex] (4) [tex]\frac{{105}}{{304}}[/tex]
Ans. The difference of [tex]1\frac{3}{{16}}[/tex] ( [tex]1\frac{3}{{16}} = \frac{{19}}{{16}}[/tex] )and its reciprocal [tex]\frac{{16}}{{19}}[/tex] is
[tex]\begin{array}{l}
\frac{{19}}{{16}} - \frac{{16}}{{19}}\\
= \frac{{{{19}^2} - {{16}^2}}}{{16 \times 19}}\\
= \frac{{\left( {19 + 16} \right)\left( {19 - 16} \right)}}{{16 \times 19}}\\
= \frac{{35 \times 3}}{{16 \times 19}}\\
= \frac{{105}}{{304}}
\end{array}[/tex]
77. Find the odd one out:
(1) Square (2) Circle (3) Parallelogram (4) Rectangle
Ans. Since Square, Parallelogram and Rectangle have four side and four angle but Circle has no side and no angle. So circle is odd one.
78. At what rate of simple interest a certain sum will be double in 15 years ?
(1) [tex]5\frac{1}{2}\% [/tex] p.a (2) 6% p.a (3) [tex]6\frac{2}{3}\% [/tex] p.a (4) [tex]7\frac{1}{2}\% [/tex] p.a
Ans. Let the rate of the interest be r% and the principle be Rs. P.
A = P + S.I
Here amount ( A ) = 2P and time = 15 years.
i.e 2P = P + S.I
P = S. I
[tex]\begin{array}{l}
P = \frac{{P \times r \times T}}{{100}}\\
\Rightarrow 1 = \frac{{r \times 15}}{{100}}\\
\Rightarrow r = \frac{{100}}{{15}} = \frac{{20}}{3} = 6\frac{2}{3}\%
\end{array}[/tex]
79. The area of a square field is [tex]6050{m^2}[/tex] . The length of its diagonal is
(1) 110 m (2) 112 m (3) 120 m (4) 135 m
Ans. Let the length of one side of square is a m . Then the area of square is [tex]{a^2}{m^2}[/tex] .
From the question [tex]{a^2} = 6050 \Rightarrow a = \sqrt {6050} = \sqrt {25 \times 2 \times 11 \times 11} = 55\sqrt 2 [/tex]
The length of the diagonal is
[tex]\begin{array}{l}
\sqrt {{a^2} + {a^2}}m \\
= \sqrt {2{a^2}} m\\
= \sqrt {2 \times {{55}^2} \times 2}m \\
= 2 \times 55m = 110m
\end{array}[/tex]
80. A tree increases annually by [tex]\frac{1}{8}[/tex] of its height. By how much will it increase after 2 years, if it stands today 64 cm high?
(1) 72 cm (2) 74 cm (3) 75 cm (4) 81 cm
Ans. After 1 years the height of tree increase [tex]\frac{1}{8} \times 64cm = 8cm[/tex] . Now after 1 year the height of tree is ( 64 + 8 ) cm = 72 cm.
Then next year its height increase [tex]\frac{1}{8} \times 72cm = 9cm[/tex] . Then the height of tree of this year is
( 72 + 9 ) cm = 81 cm.
81. If [tex]x + \frac{1}{x} = 2[/tex] ,then the value of [tex]{x^{2016}} + {x^{ - 2016}}[/tex] is
(1) 0 (2) -2 (3) 2 (4) 1
Ans.
[tex]\begin{array}{l}
x + \frac{1}{x} = 2\\
\Rightarrow {x^2} + 1 = 2x\\
\Rightarrow {x^2} - 2x + 1 = 0\\
\Rightarrow {\left( {x - 1} \right)^2} = 0\\
\Rightarrow x = 1\\
{x^{2016}} + {x^{ - 2016}}\\
= {1^{2016}} + {1^{ - 2016}}\\
= 1 + 1\\
= 2
\end{array}[/tex]
82. [tex]{\left( {997} \right)^2} + {\left( {998} \right)^2} + {\left( {999} \right)^2} - 997 \times 998 - 998 \times 999 - 999 \times 997[/tex]
(1) 0 (2) 1 (3) 2 (4) 3
Ans.
[tex]\begin{array}{l}
{\left( {997} \right)^2} + {\left( {998} \right)^2} + {\left( {999} \right)^2} - 997 \times 998 - 998 \times 999 - 999 \times 997\\
= \frac{1}{2}\left[ {2{{\left( {997} \right)}^2} + 2{{\left( {998} \right)}^2} + 2{{\left( {999} \right)}^2} - 2 \times 997 \times 998 - 2 \times 998 \times 999 - 2 \times 999 \times 997} \right]\\
= \frac{1}{2}\left[ {\left\{ {{{\left( {997} \right)}^2} - 2 \times 997 \times 998 + {{\left( {998} \right)}^2}} \right\} + \left\{ {{{\left( {998} \right)}^2} - 2 \times 998 \times 999 + {{\left( {999} \right)}^2}} \right\} + \left\{ {{{\left( {999} \right)}^2} - 2 \times 999 \times 997 + {{\left( {997} \right)}^2}} \right\}} \right]\\
= \frac{1}{2}\left[ {{{\left( {998 - 997} \right)}^2} + {{\left( {999 - 998} \right)}^2} + {{\left( {999 - 997} \right)}^2}} \right]\\
= \frac{1}{2}\left[ {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 2 \right)}^2}} \right]\\
= \frac{1}{2}\left[ {1 + 1 + 4} \right]\\
= \frac{1}{2} \times 6 = 3
\end{array}[/tex]
83. Two whole numbers whose sum is 64, cannot be in the ratio -
(1) 5 : 3 (2) 7 : 1 (3) 3 : 4 (4) 9 : 7
Ans. Option (3) is not possible because if we take the numbers as 3x and 4x .
Then 3x + 4x = 64
i.e 7x = 64
[tex]x = \frac{{64}}{7}[/tex] is not s whole number.
84. On a certain sum the simple interest at the end of [tex]12\frac{1}{2}[/tex] years becomes [tex]\frac{3}{4}[/tex] of the sum. The rate percent per annum is
(1) 4% (2) 5% (3) 6% (4) 8%
Ans. Let the principle be P [tex] \Rightarrow S.I = \frac{3}{4}P[/tex]
We know that
[tex]\begin{array}{l}
S.I = \frac{{P \times R \times T}}{{100}}\\
\Rightarrow \frac{3}{4}P = \frac{{P \times R \times 25}}{{100 \times 2}}\\
\Rightarrow \frac{3}{4} = \frac{R}{8}\\
\Rightarrow R = 6\%
\end{array}[/tex]
85. What comes next in the sequence 5.2 , 4.8 , 4.4 , 4 , _ = ?
(1) 3 (2) 3.3 (3) 3.5 (4) 3.6
Ans. 4.8 = ( 5.2 - 0.4 ) i.e 2nd term = ( 1st term - 0.4 )
4.4 = ( 4.8 - 0.4 ) i.e 3rd term = ( 2nd term - 0.4 )
4 = (4.4 - 0.4 ) i.e 4th term = ( 3rd term - 0.4 )
Then 5th term = ( 4th term - 0.4 ) or 5th term = ( 4 - 0.4 ) = 3.6
86. If the sides of a triangle are in the ratio 3 : 4 : 5 , then the largest angle is
(1) [tex]{50^0}[/tex] (2) [tex]{75^0}[/tex] (3) [tex]{90^0}[/tex] (4) [tex]{120^0}[/tex]
Ans. Let the sides of the triangle is 3x cm , 4x cm and 5x cm. We know the largest angle is belongs to the opposite site of the largest side of triangle. Now we see that
[tex]{\left( {5x} \right)^2} = {\left( {3x} \right)^2} + {\left( {4x} \right)^2}[/tex]
This is the formula of right angle triangle. Then the largest angle will be [tex]{90^0}[/tex] , because sum of three angle of a triangle is [tex]{180^0}[/tex]. One angle is [tex]{90^0}[/tex] then the sum of other two angle is [tex]{90^0}[/tex]. They always smaller than [tex]{90^0}[/tex].
87. What decimal fraction is 40 ml of a liter ?
(1) 0.4 (2) 0.04 (3) 0.004 (4) 0.0004
Ans. We know 1 l = 1000 ml
Then 1000 ml = 1 l
Therefore 40 ml = [tex]\frac{1}{{1000}} \times 40l = 0.04l[/tex]
88. In a school the ratio of boys and girls is 4 : 5 . When 100 girls leave the school, the ratio becomes 6 : 7 . How many boys are there in the school ?
(1) 1300 (2) 1500 (3) 1600 (4) 1200
Ans. Let the boys and girls are 4x and 5x respectively. When 100 girls leave the school then number of girls in the school is ( 5x - 100 ).
Now from the question 4x : ( 5x - 100 ) = 6 : 7
[tex]\begin{array}{l}
\Rightarrow \frac{{4x}}{{5x - 100}} = \frac{6}{7}\\
\Rightarrow 28x = 30x - 600\\
\Rightarrow 2x = 600\\
\Rightarrow x = 300
\end{array}[/tex]
Number of boys in the school is [tex]4x = 4 \times 300 = 1200[/tex]
89. The area of a circle is [tex]24.64{m^2}[/tex]. The circumference of the circle is
(1) 14.64 m (2) 16.36 m (3) 17.60 m (4) 18.40 m
Ans. Let the radious of circle is a m. Then it area is [tex]\pi {a^2}{m^2}[/tex].
Therefore [tex]\pi {a^2} = 24.64 \Rightarrow a = \sqrt {\frac{{24.64}}{\pi }} m[/tex]
The circumference of the circle is
[tex]\begin{array}{l}
2\pi a = 2\pi \sqrt {\frac{{24.64}}{\pi }} m\\
= 2\sqrt {\frac{{2464 \times 22}}{{100 \times 7}}} m\\
= 2\sqrt {\frac{{8 \times 4 \times 7 \times 11 \times 2 \times 11}}{{100 \times 7}}} m\\
= \frac{{2 \times 11 \times 8}}{{10}}m\\
= 17.6m
\end{array}[/tex]
90. A can cultivate [tex]\frac{2}{5}th[/tex] of the land in 6 days and B can cultivate [tex]\frac{1}{3}rd[/tex] of the land in 10 days. Working together A and B can cultivate [tex]\frac{4}{5}th[/tex] of the land in
(1) 4 days (2) 5 days (3) 8 days (4) 10 days
Ans. A's 1 days work [tex] = \frac{2}{{5 \times 6}} = \frac{1}{{15}}[/tex]
B's 1 day work [tex] = \frac{1}{{3 \times 10}} = \frac{1}{{30}}[/tex]
A and B total 1 day work
[tex]\begin{array}{l}
= \frac{1}{{15}} + \frac{1}{{30}}\\
= \frac{3}{{30}}\\
= \frac{1}{{10}}
\end{array}[/tex]
They cultivate the whole field in 10 days. They cultivate [tex]\frac{4}{5}th[/tex] of the field in [tex]\left( {\frac{4}{5} \times 10} \right)[/tex] days = 8 days
91. A container contains x kg of milk . From this container , y kg of milk was taken out and replace by water . this process was further repeated ( n - 1 ) times . is row there in the container ?
(1) [tex]x{\left( {1 - \frac{y}{x}} \right)^n}kg[/tex] (2) [tex]x{\left( {1 - \frac{y}{x}} \right)^2}kg[/tex] (3) [tex]x{\left( {1 - \frac{y}{x}} \right)^{n - 1}}kg[/tex] (4) [tex]x{\left( {1 - \frac{y}{x}} \right)^3}kg[/tex]
Ans. After 1st times container contains milk is [tex]\left( {x - y} \right)kg = x{\left( {1 - \frac{y}{x}} \right)^1}kg[/tex] .
After 2nd times container contains milk is [tex][x{\left( {1 - \frac{y}{x}} \right)^1} - y]kg = x\left( {1 - \frac{y}{x} - \frac{y}{x}} \right)kg = x\left( {1 - 2\frac{y}{x}} \right)kg[/tex].
After 3rd times container contains milk is [tex]\left[ {x\left( {1 - 2\frac{y}{x}} \right) - y} \right]kg = x\left( {1 - 2\frac{y}{x} - \frac{y}{x}} \right)kg = x\left( {1 - 3\frac{y}{x}} \right)kg[/tex]
Now after ( n - 1 ) times container contains milk is [tex]x\left( {1 - (n - 1)\frac{y}{x}} \right)kg[/tex]
Since
[tex]\begin{array}{l}
x{\left( {1 - \frac{y}{x}} \right)^{\left( {n - 1} \right)}} = x\left( {1 - \left( {n - 1} \right)\frac{y}{x} + C_2^{n - 1}\frac{{{y^2}}}{{{x^2}}} - C_3^{n - 1}\frac{{{y^3}}}{{{x^3}}} + ..............{{\left( { - 1} \right)}^{n - 1}}\frac{{{y^{n - 1}}}}{{{x^{n - 1}}}}} \right) \approx x\left( {1 - \left( {n - 1} \right)\frac{y}{x}} \right)\\
\therefore x\left( {1 - \left( {n - 1\frac{y}{x}} \right)} \right) = x{\left( {1 - \frac{y}{x}} \right)^{\left( {n - 1} \right)}}
\end{array}[/tex]
92. What number should come next in the sequence 20 , 27 , 23 , 37 , 26 , - ?
(1) 44 (2) 47 (3) 50 (4) 53
Ans. Since the difference of 20 , 23 and 26 is 3. The difference of 27 and 37 is 10. Then the next number is always be ( 37 + 10 ) = 47
93. The value of [tex]\left[ {{{\left( {0.98} \right)}^3} + {{\left( {0.02} \right)}^3} + 3 \times 0.98 \times 0.02 - 1} \right][/tex] is
(1) 1 (2) 1.09 (3) 1.98 (4) 0
Ans.
[tex]\begin{array}{l}
{\left( {0.98} \right)^3} + {\left( {0.02} \right)^3} + 3 \times 0.98 \times 0.02 - 1\\
= {\left( {1 - 0.02} \right)^3} + {\left( {0.02} \right)^3} + 3 \times \left( {1 - 0.02} \right) \times 0.02 - 1\\
= 1 - 3 \times 0.02 + 3 \times {\left( {0.02} \right)^2} - {\left( {0.02} \right)^3} + {\left( {0.02} \right)^3} + 3 \times 0.02 - 3 \times {\left( {0.02} \right)^2} - 1\\
= 0
\end{array}[/tex]
94. What number should come next in the sequence 6 , 18 , 72 , 360 , 2160 , _ ?
(1) 12120 (2) 13120 (3) 14120 (4) 15120
Ans. 18 = 6 multiply 3 or 2nd term = 1st term multiply 3
72 = 18 multiply 4 or 3rd term = 2nd term multiply 4
360 = 72 multiply 5 or 4th term = 3rd term multiply 5
2160 = 360 multiply 6 or 5th term =4th multiply 6
6th term = 5th term multiply 7 or 6th term = 2160 multiply 7= 15120
95. A student was asked to divide a number by 3, but inside dividing it , he multiplied by 3 and got 29.7. The correct answer was
(1) 3.3 (2) 9.3 (3) 9.8 (4) 9.9
Ans. Let the number is x. Then from the question we say
3x = 29.7
i.e x = 9.9
The correct ans is [tex]x \div 3 = 9.9 \div 3 = 3.3[/tex]
96. If a : b = 3 : 4 , then ( 7a + 3b ) : ( 7a - 3b ) = ?
(1) 4 : 3 (2) 5 : 2 (3) 11 : 3 (4) 37 : 19
Ans. [tex]a:b = 3:4 \Rightarrow \frac{a}{b} = \frac{3}{4} \Rightarrow a = \frac{3}{4}b[/tex]
[tex]\begin{array}{l}
\left( {7a + 3b} \right):\left( {7a - 3b} \right)\\
= \left( {7 \times \frac{3}{4}b + 3b} \right):\left( {7 \times \frac{3}{4}b - 3b} \right)\\
= \frac{1}{4}\left( {21b + 12b} \right):\frac{1}{4}\left( {21b - 12b} \right)\\
= 33b:9b\\
= 11:3
\end{array}[/tex]
97. What number should come next in the sequence 7 , 14 , 23 , 34 , 47 , _ ?
(1) 62 (2) 60 (3) 64 (4) 66
Ans. The difference between 1st term and 2nd term is ( 14 - 7 ) = 7
The difference between 2nd term and 3rd term is ( 23 - 14 ) = 9
The difference between 3rd term and 4th term is ( 34 - 23 ) = 11
The difference between 4th term and 5th term is ( 47 - 34 ) = 13
Since the difference is increased by 2 . Then the difference between 5th and 6th term will be 15. Then next term will be
( 47 + 15 ) = 62
98. Which of the following sets of numbers is in ascending order ?
(1) [tex]\frac{9}{{11}},\frac{7}{8},\frac{5}{7}[/tex] (2) [tex]\frac{7}{8},\frac{5}{7},\frac{9}{{11}}[/tex] (3) [tex]\frac{5}{7},\frac{9}{{11}},\frac{7}{8}[/tex] (4) [tex]\frac{5}{7},\frac{7}{8},\frac{9}{{11}}[/tex]
Ans. The L.C.M of 7 , 8 and 11 is 616 .
\[\begin{array}{l}
\frac{5}{7} = \frac{{5 \times 88}}{{7 \times 88}} = \frac{{440}}{{616}}\\
\frac{9}{{11}} = \frac{{9 \times 56}}{{11 \times 56}} = \frac{{504}}{{616}}\\
\frac{7}{8} = \frac{{7 \times 77}}{{8 \times 77}} = \frac{{539}}{{616}}
\end{array}\]
Since [tex]\frac{5}{7} < \frac{9}{{11}} < \frac{7}{8}[/tex]
99. The least number which when diminished by 5, is divisible by each one of 21 , 28 , 36 , 45 is
(1) 425 (2) 1259 (3) 1260 (4) 1265
Ans. The L.C.M of 21 , 28 , 36 , 45 is [tex]7 \times 3 \times 4 \times 3 \times 5 = 1260[/tex].
The least number is 1260 + 5 =1265
100. How many digits are required for numbering the pages of a book having 300 pages ?
(1) 299 (2) 492 (3) 789 (4) 792
Ans. For numbering pages 1 to 9 Number of digits = 9
pages 10 to 99 Number of digits = [tex]2 \times 90 = 180[/tex]
pages 100 to 300 Number of digits = [tex]3 \times 201 = 603[/tex]
Total no.of digit = 603 + 180 + 9 = 792
