Quant Question of the Day Chat

ok

options would help

12 is not there in options then?

oh

i must be missing something then because i got 12 and 15, let me check

could you respond to my personal message?

are you sure the answer is 15, because 15th angle will exceed 180. it has to be 12

Can someone share complete explanation?

Got it

Factorize a^2-b^ as (a+b)(a-b). a-b = 1, hence a+b = 169. Solving simultaneously, we can get both a & b. Hence both the statements together are required to answer the question.

A took a loan from B of Rs. 1040 at 5% on a condition that he would pay it in 4 yearly installments. Out of which first 3 installment would be 1/4 of the principal and the rest amount would be paid in last installment.

Find the amount paid by A in the last installment.

Let’s try to solve this problem together.👍

First, we need to find the amount of each of the first three installments. Since they are 1/4 of the principal, we can divide 1040 by 4 to get 260. So, A pays 260 rupees each year for the first three years.

Next, we need to find the amount of interest that accumulates on the loan over the four years. Since the interest rate is 5%, we can multiply 1040 by 0.05 to get 52. This is the interest for one year. To find the total interest, we need to add up the interest for each year, taking into account that the principal decreases as A pays off some of the loan.

The interest for the first year is 52 rupees, since the principal is still 1040. The interest for the second year is 52 - (260 x 0.05) = 39 rupees, since the principal is now 1040 - 260 = 780. The interest for the third year is 52 - (520 x 0.05) = 26 rupees, since the principal is now 1040 - 520 = 520. The interest for the fourth year is 52 - (780 x 0.05) = 13 rupees, since the principal is now 1040 - 780 = 260.

The total interest is 52 + 39 + 26 + 13 = 130 rupees.

Finally, we need to find the amount of the last installment. This is equal to the remaining principal plus the total interest. The remaining principal is 1040 - (260 x 3) = 260 rupees. The total interest is 130 rupees. So, the last installment is 260 + 130 = **390 rupees**.

Therefore, A pays **390 rupees** in the last installment.

Got it, Thanks

I got Confused with months,

DS Question 1 - July 26

Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party?

(1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians.

(2) 30% of the guests were vegetarian non-students.

Source: Manhattan | Difficulty: Hard

PS Question 1 - July 26

A biased dice is such that any odd number appears thrice as frequently as any even number. If the dice is rolled twice, what is the probability that the sum of the numbers thus obtained is 10?

A. 1/16
B. 5/72
C. 11/144
D. 11/72
E. 11/16

Source: eGMAT | Difficulty: Hard

This is very difficult

If the dices were not biased, the question wouldn’t have been that difficult.

The part that makes this question tricky is the fact that the dices are biased. But, if we can navigate that part, the question is a cake walk.

So, before we dive deep into the solution we must take a note of an important aspect. Had the dice not been biased the probability of each number would have been 1/6 - We all know this !

But because the dice is biased the probability if odd number appearing is three times that of an even number. We have three odd and three even numbers on the face of a dice.

If the probability of an even, P(e), number is x, the probability of an odd number, P(o), is 3x. As there are three even numbers and three odd numbers,

3(P(e)) + 3(P(o)) = 1

3(x) + 3(3x) = 1

12x = 1

x = 1/12

P(e) = 1/12

P(o) = 3/12

Once, we have got this we can apply the regular concept to solve the question.

E..
P(odd) = 3/4
P(even) = 1/4

Required cases = (6,4),(5,5) and (4,6)

Probability= 1/4*1/4 + 3/4*3/4 + 1/4*1/4 = 11/16

In a four-digit number with distinct digits, the sum of the middle digits equals the sum of the extreme digits. The sum of its second and fourth digits equals five times the sum of its other two digits. If the sum of its digits is 18, what is the sum of all the possible values of the hundreds digit?

Can anyone explain the solution ?

Consider the 4-digit number to be ’abcd’. The question is asking you to find the possible values of ’b’. Solve the equations from the statements given in the question and you will get:
1. a+c=3
2.b= 3a+2c
The combination of (a,c) will be (2,1) (3,0) (1,2). Note that a can’t be 0 because abcd is a 4-digit number.
Based on the above, b= 8 or 9 or 7. Sum of these three numbers is 24

I have my test in 15 days but I’m very weak in geometry.. It will be fine if skip geometry or practiced just 600 level questions?

Don’t skip geometry at all . I understand what u would we feeling because my geometry is very weak . I suggest you that make sure all formula and basics are clear with you as GMAT will give you combination of 2 (Geometry + another topic 650+ level questions) . and also practice atleast 650+ questions for better score.

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