MULTIPLE CHOICE QUESTIONS AND ANSWERS
1. A hall is 13 metres
53 cm. long and 8 metres 61cm. broad is to be paved with minimum number of
square tiles. The number of tiles required is:
(a) 123
(b) 77
(d) 57
(c) 99
Solution: (b)
13m 53 Icm = 135 cm and 8m 61cm = 861 cm. H.C.F. of 1353 and 861 is 123, Now
since minimum tiles are required for having the floor so area of the title has
to be the greatest so its size is greatest which is 123 cm.
Thus the number of square
tiles
1353 x 861 / 123 x 123 =
11x 7 =77
2. A shopkeeper has
three kinds of sugar 184 kg, 230 kg and 276 kg. He wants to store it into
minimum number of bags of equal size without mixing. Find the size of the bag
and the number of bags required to do the needful.
(a) 23 kg: 30
(b) 46 kg: 23
(c) 46 kg: 15
(b) 46 kg; 23
Solution: c) Size of
the bag is the H.C.F. of the numbers 184, 230, 276 which is 46. The number of
bags = 184/46 + 230/46 + 276/46 = 4 + 5 + 6 = 15
3. The greatest number
that will divide m and n m = 25 x 38 x 510 and
n = 27 x 38 x 712 is: completely where
(a) 25 x 37
x 510
(b) 25 x 37
(c) 27 x 38
x 712
(d) 27 x 37
Solution: c)
4. The greatest number
that will divide 719 and 930 leaving respectively is: remainders 5 and 6
(a) 32
(b) 38
(c) 42
(d) 97
Solution: (c) Such a
number is the H.C.F. of 719-5 and 930-6 i.e. the H.C.F. of 714 and 924 which is
42.
5. Two numbers whose
sum is 125 and their H.C.F. is 15 are:
(a) 45, 80
(b) 46, 60
(c) 45, 70
(d) They do not exist
Solution: (d) Sum of two numbers must be divisible by their
H.C.F., which is not this case so two numbers do not exist.
6. The greatest number
which when divided 1370 and 1760 by it leaves remainder 5 in each case is:
(a) 195
(b) 205
(c) 137
(d) 176
Solution: (a) To find such a number, subtract 5 from each
number and then find the H.C.F.
ð the H.C.F. of 1370-5 and 1760 – 5 is 195
7. The least number by
which 825 must be multiplied so that the resulting number may be a multiple of
715 is:
(a) 11
(b) 5
(c) 13
(d) 55
Solution: c) If n is a least number, then 825 n must be a
multiple of 715 i.e. 825/715 n is a whole number. H.C.F. of 825
and 715 is 55.
Thus 825/715 n = (15 x
55)/(13 x 55) n = 15/13n
So, n = 13.
8. There are four prime
numbers written in the ascending order of magnitude. The product of first three
numbers and that of the last three numbers is 385 and 1001 respectively. The
fourth prime number is:
(a) 11
(b) 13
(c) 17
(d) 19
Solution: (b) Let P,Q,R and s be the four prime numbers, so
that PQR = 385 and qrs = 1001
ð QR is the H.C.F. of 385 and 1001 which is 77.
ð S = 1001 + 77 = 13
9. In a school there
are 391 boys and 323 girls. These are to be divided into the largest possible
equal classes, so that there are equal number of boys and girls in each class.
How many classes are possible?
(a) 32
(b) 37
(c) 42
(d) 49
Solution: c) The number of boys and girls in each class is the
H.C.F. of 391 and 323 which is 17. Number of classes= 391/17 + 323/17 = 23 + 19
= 42
10. A man was engaged
for a certain number of days for Rs. 404.30p but being absent for some days he
was paid only rs. 279.90p. His daily wages cannot exceed by:
(a) Rs.29.10p
(b) Rs.31.30p
(c) Rs.31.10p
(d) Rs.31.41p
Solution: c) His maximum daily wages must be the H.C.F. of
404.30 and 279.90 which is Rs.31.30p
Post a Comment