Let the total number of matches to played in the tournament be ‘x’.
Given that A county cricket team has won 10 matches and lost 4.
That means total number of matches played = 14.
So,
= > 70% of x = 14
= > (70/100) * x = 14
= > 70x = 1400
= > x = 20.
So, the number of matches to be played = 20.
They have a record of exactly 75% wins = 20 * 75/100
= > 15.
Therefore, the number of matches should the team win = 15 – 10 = 5.
Hope this helps!
(b) 16.66%
Earlier for ₹x we could purchase y gm of sugar.
Now we pay ₹1.2x for y gm of sugar
(As there was an increase in price so, x + 20%x = 1.2x)
At current rates for ₹x you can purchase y/1.2 gm of sugar
So the reduced consumption is y-(y/1.2)
Percentage change = (reduced consumption/ original consumption ) *100
That is (0.2/1.2) *100 = 16.66% (approx)
3606
My standard for success is to build confidence, be able to earn money, and meet people.
total no of 4 wheelers: 40
total no of 2 wheelers: 18
3500*10/100
3500*11.5/100 = 402.5 per year
402.5×3 = R1207.5
Let the number of one rupee coins in the bag be x.
Number of 50 paise coins in the bag is 93 – x.
Total value of coins
[100x + 50(93 – x)]paise = 5600 paise
=> x = 74
ANS = 74
64
The number of ways of selecting a group of eight is
5 men and 3 women=5C5×6C3 =20
4 men and 4 women=5C4×6C4 =75
3 men and 5 women=5C3×6C5=60
2 men and 6 women=5C2×6C6=10
Thus the total possible cases is 20+75+60+10=165.
Answer: 8:10, 7:10
Explanation:
The bus b1, which started at P, reached S at 10:40, passing through the intermediary cities Q and R.
The time taken to travel from P to S
= 3 * 40 + 2 * 15 = 150minute
(journey)+(stoppage) = 2 hrs 30 minutes.
Hence, b1 started at 10:40 – 2:30 = 8:10 at P.
b2 reached Q, starting at U, through the city T, S and R.
The time taken by it to reach S = 4 * 40 + 3 * 15 = 205 minutes = 3 hr 25 minutes.
Hence, b2 started at, 10:35 – 3:25 = 7:10, at U.
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
31
1*2=2
2*2=4
2*4=8
8*4=32
Ans : 32