A
46
225
If Vijay gives ‘x’ marbles to Ajay then Vijay and Ajay would have V – x and A + x marbles.
V – x = A + x — (1)
If Ajay gives 2x marbles to Vijay then Ajay and Vijay would have A – 2x and V + 2x marbles.
V + 2x – (A – 2x) = 30 => V – A + 4x = 30 — (2)
From (1) we have V – A = 2x
Substituting V – A = 2x in (2)
6x = 30 => x = 5.
Question is not completed
The letters A, B, C, D, E, F and G, not necessarily in that order, stand for seven consecutive integers from 1 to 10
D is 3 less than A
B is the middle term
F is as much less than B as C is greater than D
G is greater than F
1. The fifth integer is
(a) A
(b) C
(c) D
(d) E
(e) F
ans:a
2.A is as much greater than F as which integer is
less than G
(a) A
(b) B
(c) C
(d) D
(e) E
ans:a
3. If A = 7, the sum of E and G is
(a) 8
(b) 10
(c) 12
(d) 14
(e) 16
4. An integer T is as much greater than C as C is
greater than E. T
can be written as A + E. What is D?
(a) 2
(b) 3
(c) 4
(d) 5
(e) Cannot be determined
ans:a
X-Y=-1
40
Answer is 45
First we need to subtract those reminders from the respective numbers, then we have to find the hcf of two numbers(numbers got from the subtraction) then you will get the answer.
So,
After subtraction you will get
3026-11 = 3015
5053-13 = 5040
HCF of these two numbers
5 | 3015 5040
3 | 603 1008
3 | 201 336
| 67 112
We can’t find a common diviser since 67 is a prime number
So the HCF = 5 * 3 * 3
= 45
5*3*3 = 45
A’s speed=6 mph ,B’s speed=8 mph
Let, after x hrs, they will meet.
so, the distance traveled by A in x hrs should be the same as the distance traveled by B in (x-1/2)hrs [as B started the journey after 30 min of A]
Thus, 6x=8(x-1/2)[as distance=speed*time]
=>8x-6x=4
=>2x=4
=>x=2
after 2 hrs they will meet so time=(9+2)=11.00 a.m
99/1020
16 chapatis* 6 Rs= 96
5 plates of rice * 45 Rs = 225
7 plates of mixed vegetable * 70 Rs = 490
6 ice-cream * ??
The amount that Alok paid the cashier was Rs.961
96+225+490 = 811
961-811= 150
150/6 = 25
The cost of each ice-cream cup is 25 .Rs
let present age of ANAND AND BALA BE( A AND B ) RESPECTIVELY
A.T.Q
(A-10) = 1/3[B-10]. ……… – {1}
given
B = A+ 12 PUTTING IN 1
THEREFORE A-10 = 1/3 (A+2) = A+2 = 3A – 30
2A =32
A=16
ans: none
let it may be any number the square cant end in 8
21 days