Mathematics Test
1. The L.C.M. (Lowest (or Least) Common Multiple) of two numbers is 45 times to their H.C.F (Highest Common Factor). If one of the numbers is 125 and sum of L.C.M. and H.C.F. is 1150, the other number is:
a) 215 b) 220
c) 225 d) 235
2. Six bells commence tolling together and toll at intervals 2,4,6,8,10 and 12 seconds respectively. In 30 minutes how many times they toll together.
a) 4 b) 10
c) 15 d) 16
3. The value of (0.625 * 0.0729 * 28.9)/(0.0017 * 0.025 * 8.1) is
a) 0.3825 b) 3.825
c) 38.25 d) 3825
4. The number of coins 1.5 cm in diameter and 0.2cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm is:
a) 380 b) 450
c) 472 d) 540
5. The size of a wooden block is 5 * 10 * 20 cm3. How many whole such blocks you will take to construct a solid wooden cube of minimum size?
a) 6 b) 8
c) 12 d) 16
6. Find the odd man out - 1050, 510, 242, 106, 46, 16, 3
a) 510 b) 242
c) 106 d) 46
7. If x and y are the two digits f the number 653xy such that this number is divisible by 80, then x+y is equal to:
a) 2 b) 3
c) 4 d) 6
8. If 1.5x=0.04y then the value of (y-x)/(y+x) is
a) 730/77 b) 73/77
c) 7.3/77 d) None
9. The average age of a class is 15.8 years. The average age of boys in the class is 16.4 years while that of girls is 15.4 years. What is the ratio of boys to girls in the class?
a) 1:2 b) 3:4
c) 3:5 d) None of these
10. If one-seventh of a number exceeds its eleventh part by 100 then the number is…
a) 770
b) 1100
c) 1825
d) 1925
11. The ratio of Darsh's age to her mother's age is 3:8. The difference of their ages is 35 years. The ratio of their ages after 4 years will be:
a) 7:12 b) 5:12
c) 38:43 d) 42:47
12. Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what ratio should these be mixed to get an alloy 15 times as heavy as water?
a) 1:1 b) 2:3
c) 1:2 d) 3:2
Answer
1. c Let L.C.M. be l and H.C.F. be h. Then l = 45h
l + h =1150
This gives h =25 and l = 1125
Thus the second number = (25 * 1125)/125 = 225
2. d Solution: L.C.M. of 2,4,6,8,10,12 = 120
So they toll together after every 120 seconds i.e. 2 minutes. So in 30 minutes they toll together 30/2 + 1 = 16 times
3. d Solution: The sum of decimal places in numerator and denominator being the same, decimal point can be removed
(625*729*289) / (17*25*81) = 3825
4. b Solution: Volume of 1 coin = 22/7 *(1.5)/2*(1.5)/2 * 0.2 = 99/280 cm3
Volume of large cylinder = 22/7 *(4.5)/2*(4.5)/2 * 10 = (99*45)/28cm3
Number of coins = (99*45)/28 * 280/99 = 450
5. b Solution: Size of each cube = 5 cm
So number of blocks = (5 * 10 * 20)/(5*5*5) = 8
6. c Solution:
2nd term = (Ist term - 30) / 2 = (1050-30)/2 = 510
3rd term = (2nd term - 26) / 2 = (510-26)/2 = 242
4th term = (3rd term - 22) / 2 = (242-22)/2 = 110
So 106 is the answer.
7. a Solution:
Since 653xy is divisible by 2 as well as by 5, so y = 0
Now 653x0 is divisible by 8 so 3x0 is also divisible by 8.
By hit and trial x=2 and x+y = 2
8. b Solution: x/y = 0.04/1.5 = 2/75
So (y-x)/(y+x) = (1 - x/y)/(1 + x/y) = (1 - 2/75)/ (1 + 2/75) = 73/77
9. d Solution:
Let the ratio be k:1.
Then k*16.4+1*15.4 + (k+1)*15.8 (16.4-15.8)k = 15.8 - 15.4 k=0.4/0.6 = 2/3 so required ratio = 2:3
10. d Solution:
Let the number be x. Then X/7 - x/11 =100 11x-7x = 7700 x=1925.
11. b Solution:
Let their ages be 3x and 8x
8x - 3x =35
x =7
Their present ages are 21 and 56 years.
Ratio of their ages after 4 years are 25:6 = 5:12
12. d Solution:
Let 1gm of gold be mixed with x gm of copper to give (1+x)gm of the alloy.
1G=19W, 1C = 9W and alloy = 15W 1gm gold + xgm Copper = (1+x)gm alloy 19W+9Wx = (1+x)*15W x = 4W/6W = 2/3
So ratio of gold and copper is 1:2/3 or 3:2
{2006/08/09}
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