Quantitative Aptitude
1)4003 * 77 - 21015 = ? * 116
a)2477
b)2478
c)2467
d)2476
e)none of these
Sol> Suppose unknown value is x
then
4003 * 77 - 21015 = x * 116
=> (4003 * 77 -21015 )/116 = x
=>(308231 -21015 )/116 = x
=> (287216)/116 =x
=>x=2476
2)[( 5 √7 + √7 ) * (4√7 + 8√7)] - (19)^2 = ?
a)143
b)72√7
c)134
d)70√7
e)none of these
Sol>
[ (5 √7 + √7 ) * (4√7 + 8√7)] - (19)^2
=[6√7 * 12 √7 ] - 361
= 72 √7 √7 - 361
=72 * 7 - 361
=504 - 361
=143
3)(4444 / 40) + (645 / 25) + (3991 / 26) = ?
a)280.4
b)290.4
c)295.4
d)285.4
e)none of these
Sol>
(4444 / 40) + (645 / 25) + (3991 / 26)
=>(1111/10) + ( 129/5) + (307 / 2 )
=>(1111 + 258 + 1535) /10
=>2904/10
=>209.4
4)√33124 * √2601 - (83)^2 = (?)^2 + (37) ^ 2
a)37
b)33
c)34
d)28
e)none of these
Sol> Suppose unknown value is x
√33124 * √2601 - (83)^2 = (x)^2 + (37) ^ 2
=>182 * 51 - 6889 = (x)^2 + (37) ^ 2
=> 9282 - 6889 = (x)^2 + (37) ^ 2
=> 2393 = (x)^2 + (37) ^ 2
=> 2393 =(x)^2 +1369
=>1024 =(x)^2
=> x=√1024
=>x=32
5) 5 17/37 * 4 51/52 * 11 1/7 + 2 3/4
a)303.75
b)305.75
c)303 3/4
d)305 1/4
e)none of these
Sol>
5 17/37 * 4 51/52 * 11 1/7 + 2 3/4
= 202/37 * 259 /52 * 78 /7 + 11/4
=303 + 11/4
=305.75
6 -10 What approximate value come in place of ? in the following questions
(Not need of exact value)
6) 8737 / 343 * √50 = ?
a)250
b)140
c)180
d)100
d)280
Sol>
8737 / 343 * √50
=>
8737 / 343 = 25.4 (nearly)
√50 =7.0 (nearly)
so 8737 / 343 * √50 =177.8 (nearly)
so ans from above options is 180
7) 54821^ 1/3 * (303/8) =?^2
a)48
b)38
c)28
d)18
e)58
Sol>
54821^ 1/3 = 37. (nearly)
303/8 = 37.8 (nearly)
Suppose unknown value is x
54821^ 1/3 * (303/8) =x^2
=> 37. * 37.8 =x^2
so ans is 37. nearly 38
so ans is 38
8) 5/8 of 4011.33 + 7/10 of 3411.22 = ?
a)4810
b)4980
c)4890
d)4930
e)4850
Sol>
2507.0 + 2387.8
=4894
So ans nearly 4890
9)23% of 6783 + 57 % 8431 = ?
a)6460
b)6420
c)6320
d)6630
e)6360
Sol>
1560.0 + 4805.6 (nearly)
= 6365 (nearly)
so nearest answer from above options is
6360
10) 335.01 * 244.99 / 55 = ?
a)1490
b)1550
c)1420
d)1590
e)1400
Sol>
335.01 * 244.99 / 55
=335.01 * 4.4 (nearly)
=1492 (nearly)
so nearest answer from above options is
1490
in questions a number series is given.In each series only one number is wrong.Find out wrong number.
11)5531 5506 5425 5304 5135 4910 4621
a)5531
b)5435
c)4621
d)5135
e)5506
Sol>
5531 - 5506 = 25 square root is 5
5506 - 5425 =81 square root is 9
5425 - 5304 =121 square root is 11
5304 - 5135 =169 square root is 13
5135 - 4910 =225 square root is 15
4910 -4621 =289 square root is 17
difference of square roots are same is 2 except first two terms
so the first term is wrong in series
that is 5531
12)A certain amount was to be distributed among A,B, and C in the ratio 2:3:4 respectively, but was erroneously distributed in the ratio 7:2:5 respectively. As a result of this, B got Rs.40 less, What is the amount?
a)Rs 210/-
b)Rs.270/-
c)Rs.230/-
d)Rs.280/-
e) None of these
Sol>
Suppose certain amount is x
then True distribution is (Respectively A,B,C)
(2/9)x , (3/9)x , (4/9)x
and erroneously distribution is (Respectively A,B,C)
(7/ 14)x , (2/ 14)x, (5/ 14)x
Now, B got Rs.40 less (given)
so
(3/9)x - (2/ 14)x = 40
=>(42x - 18x)/126 = 40
=>24x = 40 * 126
=>x=210
So certain amount is Rs. 210/-
13 )Rachita enters a shop to buy ice-creams ,cookies and pastries. She has to buy at least 9 unit of each. She buys more cookies than ice-creams and more pastries than cookies. She picks up a total of 32 items.
How many cookies does she buy?
a)Either 12 or 13
b) Either 11 or 12
c)Either10 or 11
d)Either 9 or 11
e)Either 9 or10
Sol>
For solving this question, I recommend you to go through options
here,
No of Pastries > No of Cookies > No of Ice creams
and has to buy at least 9 unit of each.
So suppose Ice creams for 9 units, then No of cookies must >9 so option d) and e) is not possible
Now Suppose No of Cookies is Either10 or 11 then No of Pastries must >11 , so its smallest value becomes 13
then sum of three = 9 +11+ 12= 32 , which is correct
(If we take cookies for Either 11 or 12 then Pastries>12 then sum is=9+12+13=34 not correct
If we take cookies for Either 12 or 13 then Pastries>13 then sum is=9+13+14=36 not correct)
14)The product of three consecutive even numbers is 4032. The product of the first and the third number is 252.What is the five times the second number?
a)80
b)100
c)60
d)70
e)90
Sol>
Suppose three consecutive even numbers are x-2 , x, x+2
The product of the first and the third number is 252 so
(x-2)(x+2)=252
=>x^2 - 4 =252
=>x^2 = 256
=> x=√256
=>x=16
which is second number,
five times of it is =16*5 = 80
15)The sum of the ages of 4 members of a family 5 years ago was 94 years . Today, when the daughter has been married off and replaced by a daughter in law , the sum of their ages is 92 .Assuming that there has been no other chance in the family structure and all the people are alive, what is the difference in the age of the daughter and the daughter in law?
a)22 years
b)11 years
c)25years
d)19years
e)15years
Sol>
The sum of the ages of 4 members of a family 5 years ago was 94 years
so their sum of recent ages = 94 + 4(5)
=114
Now the daughter has been married off and replaced by a daughter in law,the sum of their ages is 92
so
sum of 4 members ages - age of daughter + daughter in law 's age = 92
=>114 - age of daughter + daughter in law 's age = 92
=>114-92 = age of daughter - daughter in law 's age
=> 22= age of daughter - daughter in law 's age
So,the difference in the age of the daughter and the daughter in law = 22 years
16) Akash scored 73 marks in subject A. He scored 56% marks in subject B and X marks in subject C. Maximum marks in each subject were 150.The overall percentage marks obtained by Akash in all the three subjects together were 54%. How many mark did he score in subject C?
a)84
b)86
c)79
d)73
e)none of these
Sol>
Mark for subject A=73
Mark for subject B=56% of 150 =150*56 / 100 =84
Mark for subject C=X
Now overall percentage marks = 54%
so
(73 + 84 +X ) /450 * 100 =54
=>(73 + 84 +X )=243
=>73 + 84 +X=243
=> 157+X=243
=>X=243-157
=>X=86
Marks scored in C=86
17) The area of a square is 1444 square meters. The breadth of a rectangle is 1/4 th the side of the square and the length of the rectangle is thrice the breadth. What is the difference between the area of the square and the area of the rectangle?
a)1552.38 sq.meter
b)1169.33 sq.meter
c)1181.21 sq.meter
d)1173.25 sq.meter
e)None of these
Sol>
The area of a square is 1444 square meters
Now area of square =(side)^2
=>1444= (side)^2
=> √1444=side
=>side=38
Now breadth of a rectangle is 1/4 th the side of the square
So breadth of a rectangle = 1/4 *38
length of the rectangle is thrice the breadth = 3 (1/4 *38)
So Area of the rectangle = length * breadth
= 3 (1/4 *38) * ( 1/4 *38)
= 270.75 square meters
Now,the difference between the area of the square and the area of the rectangle
=1444 - 270. 75
=1173.25 square meters
.
Thank sir 🙏🙏
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