Test Series - w Quants

Test Number 18/24

Q: In Maa Yatri Temple every devotee offers fruits to the orphans. Thus every orphan receives bananas, oranges and grapes in the ratio of 3:2:7 in terms of dozens. But the weight of a grape is 24 gm and  weight of a banana and an orange are in the ratio of 4:5, while the weight of an orangeis 150gm. Find the ratio of all the three fruits in terms of weight, that an orphan gets
A. 180:150:82
B. 75:42:90
C. 90:75:42
D. None of these
Solution: Ratio of fruits (by dozen) = 3:2:7
Ratio of fruits (by weight) =  120 :150: 24
Therefore, Ratio of fruits (combined) by weight = 3 x 120 : 2 x 150 : 7 x 24 = 30 : 25 : 14
Q: Seats for Mathematics, Physics and Biology in a school are in the ratio 5:7:8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats ?
A. 1:2:3
B. 3:4:5
C. 2:3:4
D. 4:5:6
Solution: Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively. Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x). ⇒ [(140/100) × 5x],[(150/100) × 7x] and [(175/100) × 8x] ⇒ 7x, 21x/2 and 14x. ⇒ The required ratio =7x : 21x/2 : 14x ⇒ 14x : 21x : 28x ⇒ 2 : 3 : 4
Q: Amit is five years older than Vaibhav at present. After four years the ratio of their age will be 5:4. What is Amit age at present (in years)?
A. 21
B. 16
C. 12
D. 18
Solution: 21
Q: If Raj was one-third as old as Rahim 5 years back and Raj is 17 years old now, How old is Rahim now?
A. 48
B. 41
C. 36
D. 40
Solution: Raj’s age today = 17 decades, Hence, 5 decades back, he must be 12 years old. Rahim must be 36 years old, Because (3×12). 5 years back Rahim must be 41 years old today. Because (36+5).
Q: Satheesh is two years older than Goutham who is twice as old as Sai. If the total of the ages of Satheesh, Goutham and Sai is 27, then how old is Goutham?
A. 12
B. 11
C. 10
D. 13
Solution: 10
Q: H.C.F of 3240, 3600 and a third number is 36 and their L.C.M is 24 x 35 x 52 x 72 . What is the third number?
A. 24157
B. 42146
C. 49874
D. 47628
Solution: 47628
Q: If the sum of two numbers is 55 and the H.C.F and L.C.M of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to
A. 601/55
B. 11/120
C. 120/11
D. 55/601
Solution: Let the numbers be a and b . Then, a+b =55 and ab = 5 x 120 = 600.
 
Therefore, Required sum = 1a+1b=a+bab=55600=11120
Q: Three different containers contain 1365 litres, 1560 litres and 1755 litres of mixtures of milk and water respectively. What biggest measure that can measure all the different quantities exactly ?
A. 129 lit
B. 195 lit
C. 97 lit
D. 37 lit
Solution: From the given data,
containers contain mixtures of milk and water as 1365 lit, 1560 lit and 1755 lit.
 
Biggest measure that can measure all these different quantities exactly can be given by
HCF of (1365, 1560, 1755) 
That can be found as 
 
HCF of 1365, 1755 = 195
Now, HCF of 195 and 1560 = 195
Hence, biggest measure that can measure all the given different quantities exactly is 195 lit.
 
Q: The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
A. 127
B. 235
C. 123
D. 305
Solution: Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.

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