Find the sum of prime numbers lying between 50 to 80?
460
490
520
530
Answer: (c) 520Solution: Prime numbers lying between 50 to 80 are 53, 57, 59, 61, 67, 71, 73, 79.Now the sum of these numbers $$ 53 + 57 + 59 + 61 + 67 + 71 + 73 + 79 $$ $$ = 520 $$ Hence the sum of prime numbers lying between 50 to 80 is 520.
There are four prime numbers written in ascending order. The product of the first three prime numbers is 230 and that of the last three is 530. Find the last prime number?
47
53
57
61
Answer: (b) 53Solution: Let four prime numbers are A, B, C, and D then according to the question $$ A \times B \times C = 230.....(1) $$ $$ B \times C \times D = 530.....(2) $$ By dividing the equation (2) from equation (1) $$ \frac{B \times C \times D}{A \times B \times C} = \frac{530}{230} $$ $$ \frac{D}{A} = \frac{53}{23} $$ Hence the last number is 53.
The sum of the first 20 natural numbers is?
210
220
225
230
Answer: (a) 210Solution: The sum of n natural numbers $$ = \frac{n \ (n + 1)}{2} $$ The sum of 20 natural numbers $$ = \frac{20 \ (20 + 1)}{2} $$ $$ = \frac{20 \times 21}{2} $$ $$ = 210 $$
In a sum involving division, the divisor is 10 times the quotient and 4 times the remainder. If the remainder is 30 then find the dividend?
1400
1420
1450
1470
Answer: (d) 1470Solution: According to the question $$ divisor = 10 \times quotient....(1) $$ $$ divisor = 4 \times remainder $$ $$ divisor = 4 \times 30 $$ $$ divisor = 120 $$ by putting the value of divisor in equation (1) $$ 120 = 10 \times quotient $$ $$ quotient = \frac{120}{10} = 12 $$ Now $$ dividend = divisor \times quotient \\ + remainder $$ $$ = 120 \times 12 + 30 $$ $$ = 1440 + 30 $$ $$ dividend = 1470 $$
If a number when divided by the sum of 220 and 120 gives two times their difference as quotient and 25 as the remainder then find the number?