Q: If the amount received at the end of 2nd and 3rd year at Compound Interest on a certain Principal is Rs 33708, and Rs 35730.48 respectively, what is the rate of interest?
Solution: 6 percen
Q: Sivagami is 2 years elder than Meena. After 6 years the total of their ages will be 7 times of their current age. Then age of Sivagami is :
Solution: Let Meena’s age = A. Then Sivagami’s age = A + 2 After 6 years the total of their ages will be 7 times of what? Not clear. So the given data are inadequate.
Q: Mohan was married 5 years ago. At present his age is 6/5 times of his age at the time of marriage. His wife is 3 year younger to him. Find his wife’s present age?
Solution: Let Mohan present age = p ATQ, 5/6 x p = p - 5 p = 30 His wifes age = 30 - 3 = 27 yrs.
Q: In a joint family, there are 24 members whose age is decreased by 4 months, when one girl aged 24 years is replaced by a new boy. Find the age of the new boy ?
Solution: Age of new boy = (Age of moved girl) – (Number of family members × Decreased in average) = 24 – 24×4/12 = 24 – 8 = 16 years.
Q: A sum of Rs.312 was divided among 100 boys and girls in such a way that the boy gets Rs.3.60 and each girl Rs. 2.40 the number of girls is
Solution: Step (i): Let x be the number of boys and y be the number of girls. Given total number of boys and girls = 100 x+y=100 -------------- (i) Step (ii): A boy gets Rs. 3.60 and a girl gets Rs. 2.40 The amount given to 100 boys and girls = Rs. 312 3.60x + 2.40y = 312 -------------- (ii) Step (iii): Solving (i) and (ii) 3.60x + 3.60y = 360 --------- Multiply (i) by 3.603.60x + 2.40y = 312 --------- (ii) 1.20y = 48 y = 48 / 1.20 = 40 Number of girls = 40
Q: In a fort, there are 1200 soldiers. If each soldier consumes 3 kg per day, the provisions available in the fort will last for 30 days. If some more soldiers join, the provisions available will last for 25 days given each soldier consumes 2.5 kg per day. Find the number of soldiers joining the fort in that case ?
Solution: Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)= (1200)(30)(3) kg. Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k As the same provisions are available=> (1200)(30)(3) = (1200 + x)(25)(2.5) x = ([(1200)(30)(3)] / (25)(2.5)) - 1200 => x = 528.
Q: The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
Solution: L.C.M. of 6, 9, 15 and 18 is 90. Let required number be 90k + 4, which is multiple of 7. Least value of k for which (90k + 4) is divisible by 7 is k = 4. =>Required number = (90 x 4) + 4 = 364.
Q: Three numbers are in the ratio of 3:5:10 and their LCM is 630. Find their HCF.
Solution: 21
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