{"id":39757,"date":"2019-12-27T15:56:13","date_gmt":"2019-12-27T10:26:13","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=39757"},"modified":"2019-12-27T15:56:13","modified_gmt":"2019-12-27T10:26:13","slug":"quant-questions-for-xat-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/quant-questions-for-xat-set-2-pdf\/","title":{"rendered":"Quant Questions for XAT Set-2 PDF"},"content":{"rendered":"

Quant Questions for XAT Set-2 PDF<\/strong><\/span><\/h2>\n

Download important Quant Questions Set-2 for XAT PDF based on previously asked questions in the XAT exam. Practice Quant Questions\u00a0 Set-2 PDF for the XAT exam.<\/p>\n

Download Quant Questions PDF<\/a><\/p>\n

Get 5 XAT mocks for Rs. 299. Enroll here<\/a><\/p>\n

Question 1:\u00a0<\/b>If R = $(30^{65}-29^{65})\/(30^{64}+29^{64})$ ,then<\/p>\n

a)\u00a0$0<R\\leq0.1$<\/p>\n

b)\u00a0$0.1<R\\leq0.5$<\/p>\n

c)\u00a0$0.5<R\\leq1.0$<\/p>\n

d)\u00a0$R>1.0$<\/p>\n

Question 2:\u00a0<\/b>If x = $(16^3 + 17^3+ 18^3+ 19^3 )$, then x divided by 70 leaves a remainder of<\/p>\n

a)\u00a00<\/p>\n

b)\u00a01<\/p>\n

c)\u00a069<\/p>\n

d)\u00a035<\/p>\n

Instructions<\/b><\/p>\n

Directions for the following two questions:<\/p>\n

Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is $240 + bx + cx^2$ , where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.67%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.<\/p>\n

Question 3:\u00a0<\/b>What is the maximum daily profit, in rupees, that Mr. David can realize from his business?<\/p>\n

a)\u00a0620<\/p>\n

b)\u00a0920<\/p>\n

c)\u00a0840<\/p>\n

d)\u00a0760<\/p>\n

e)\u00a0Cannot be determined<\/p>\n

Question 4:\u00a0<\/b>Five horses, Red, White, Grey, Black and Spotted participated in a race. As per the rules of the race, the persons betting on the winning horse get four times the bet amount and those betting on the horse that came in second get thrice the bet amount. Moreover, the bet amount is returned to those betting on the horse that came in third, and the rest lose the bet amount. Raju bets Rs. 3000, Rs. 2000 and Rs. 1000 on Red, White and Black horses respectively and ends up with no profit and no loss.
\nSuppose, in addition, it is known that Grey came in fourth. Then which of the following cannot be true?<\/p>\n

a)\u00a0Spotted came in first<\/p>\n

b)\u00a0Red finished last<\/p>\n

c)\u00a0White came in second<\/p>\n

d)\u00a0Black came in second<\/p>\n

e)\u00a0There was one horse between Black and White<\/p>\n

Question 5:\u00a0<\/b>John borrowed Rs. 2,10,000 from a bank at an interest rate of 10% per annum, compounded annually. The loan was repaid in two equal instalments, the first after one year and the second after another year. The first instalment was interest of one year plus part of the principal amount, while the second was the rest of the principal amount plus due interest thereon. Then each instalment, in Rs., is<\/p>\n

XAT Solved Previous papers<\/a><\/p>
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Question 6:\u00a0<\/b>The average of 7 consecutive numbers is P. If the next three numbers are also added, the average shall<\/p>\n

a)\u00a0remain unchanged<\/p>\n

b)\u00a0increase by 1<\/p>\n

c)\u00a0increase by 1.5<\/p>\n

d)\u00a0increase by 2<\/p>\n

Question 7:\u00a0<\/b>The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is<\/p>\n

a)\u00a03 : 10<\/p>\n

b)\u00a01 : 3<\/p>\n

c)\u00a01 : 4<\/p>\n

d)\u00a02 : 5<\/p>\n

Question 8:\u00a0<\/b>If $xy + yz + zx = 0$, then $(x + y + z)^2$ equals<\/p>\n

a)\u00a0$(x + y)^2 + xz$<\/p>\n

b)\u00a0$(x + z)^2 + xy$<\/p>\n

c)\u00a0$x^2 + y^2 + z^2$<\/p>\n

d)\u00a0$2(xy + yz + xz)$<\/p>\n

Question 9:\u00a0<\/b>If $x+1=x^{2}$ and $x>0$, then $2x^{4}$\u00a0 is<\/p>\n

a)\u00a0$6+4\\sqrt{5}$<\/p>\n

b)\u00a0$3+3\\sqrt{5}$<\/p>\n

c)\u00a0$5+3\\sqrt{5}$<\/p>\n

d)\u00a0$7+3\\sqrt{5}$<\/p>\n

Question 10:\u00a0<\/b>The product of the distinct roots of $\\mid x^2 – x – 6 \\mid = x + 2$ is<\/p>\n

a)\u00a0\u221216<\/p>\n

b)\u00a0-4<\/p>\n

c)\u00a0-24<\/p>\n

d)\u00a0-8<\/p>\n

XAT Decision making practice questions<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

$\\frac{(30^{65}-29^{65})}{(30^{64}+29^{64})} = ((30-29)*\\frac{(30^{64}+30^{63}*29+….+29^{64})}{(30^{64}+29^{64})}$ , which is greater than 1 . Hence option D.<\/p>\n

2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

We know that x = $16^3 + 17^3 + 18^3 + 19^3 = (16^3 + 19^3) + (17^3 + 18^3)$<\/p>\n

= $(16 + 19)(16^2 – 16 * 19 + 19^2) + (17 + 18)(17^2 – 17 * 18 + 18^2)$ = 35 \u00d7 odd + 35 \u00d7 odd = 35 \u00d7 even = 35 \u00d7 (2k)<\/p>\n

=>\u00a0x = 70k<\/p>\n

=> Remainder when divided by 70 is 0.<\/p>\n

3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Cost of 20 units = 240+20b+400c
\nCost of 40 units = 240+40b+1600c = 5\/3 * (240+20b+400c) => 720+120b+4800c = 1200+100b+2000c
\n=> 480 = 20b + 2800c => 120 = 5b + 700c
\nCost of 60 units = 240+60b+3600c = 3\/2 (240+40b+1600c) => 480 + 120b + 7200c = 720 + 120b + 4800c
\n=> 240 = 2400c => c = 1\/10 and b = 10
\nLet the number of items needed for max profit be k
\nCP = $240+10k+k^2\/10$
\nSP = 30k
\nProfit = SP – CP = $30k – 240 – 10k – k^2\/10$ = $20k – 240 – k^2\/10$
\nMaximum when 20 – k\/5 = 0 or k = 100
\nProfit = 2000 – 240 – 1000 = 760<\/p>\n

4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

There are total 3 cases which satisfies the condition “no profit and no loss.”<\/p>\n

Case 1: White comes 2nd.(remaining two horses(red\/black) come 4th\/5th)<\/p>\n

Profit from white horse = Final Amount – Initial Amount = 2000*3 – 2000 = 4000<\/p>\n

Loss from Red and Black horse = 3000+1000 = 4000<\/p>\n

Net profit = 4000-4000 = 0<\/p>\n

Case 2: Black, Red come second, third respectively.(remaining one horse(white) comes 4th\/5th)<\/p>\n

Profit from Black = 1000*3-1000 = 2000<\/p>\n

Profit from Red = 3000 – 3000 = 0<\/p>\n

Loss from white = 2000<\/p>\n

Net profit = 2000-2000 = 0<\/p>\n

Case 3: black, white come first, third respectively.(remaining one horse(red) comes 4th\/5th)<\/p>\n

Profit from Black = 1000*4-1000 = 3000<\/p>\n

Profit from White = 2000 – 2000 = 0<\/p>\n

Loss from Red = 3000<\/p>\n

Net Profit = 3000-3000=0<\/p>\n

And it is mentioned that grey case 4th. ==> case 1 is wrong.(because, in that case red, black should come 4th,5th)<\/p>\n

So option C cannot be true.<\/p>\n

5)\u00a0Answer:\u00a0121000<\/b><\/p>\n

We have to equate the installments and the amount due either at the time of borrowing or at the time when the entire loan is repaid. Let us bring all values to the time frame in which all the dues get settled, i.e, by the end of 2 years.<\/p>\n

John borrowed Rs. 2,10,000 from the bank at 10% per annum. This loan will amount to 2,10,000*1.1*1.1 = Rs.2,54,100 by the end of 2 years.
\nLet the amount paid as installment every year be Rs.x.<\/p>\n

John would pay the first installment by the end of the first year. Therefore, we have to calculate the interest on this amount from the end of the first year to the end of the second year. The loan will get settled the moment the second installment is paid.<\/p>\n

=> 1.1x + x = 2,54,100
\n2.1x = 2,54,100
\n=> x = Rs. 1,21,000.<\/p>\n

Therefore, 121000 is the correct answer.<\/p>\n

XAT Quant formulas PDF<\/a><\/p>
\n

Take XAT Mock Test<\/a><\/p><\/p>\n

6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let the 7 consecutive numbers be a-3, a-2, a-1, a, a+1, a+2 and a+3.
\nSum of the numbers = 7a and the average of these numbers = a
\nIf next 3 numbers a+4, 4+5 and a+6 are also added then the average of these 10\u00a0 numbers = $\\dfrac{7a+a+4+a+5+a+6}{10} = a+1.5$
\nThus, the average increases by 1.5
\nHence, option C is the correct answer.<\/p>\n

7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let ‘a’, ‘b’ and ‘c’ be the concentration of salt in solutions A, B and C respectively.<\/p>\n

It is given that\u00a0three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.<\/p>\n

$\\Rightarrow$ $\\dfrac{a+2b+3c}{1+2+3} = 20$<\/p>\n

$\\Rightarrow$ $a+2b+3c = 120$ … (1)<\/p>\n

If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.<\/p>\n

$\\Rightarrow$ $\\dfrac{3a+2b+c}{1+2+3} = 30$<\/p>\n

$\\Rightarrow$\u00a0$3a+2b+c = 180$ … (2)<\/p>\n

From equation (1) and (2), we can say that<\/p>\n

$\\Rightarrow$ $b+2c = 45$<\/p>\n

$\\Rightarrow$ $b = 45 – 2c$<\/p>\n

Also, on subtracting (1) from (2), we get<\/p>\n

$a – c = 30$<\/p>\n

$\\Rightarrow$ $a = 30 + c$<\/p>\n

In solution D, B and C are mixed in the ratio 2 : 7<\/p>\n

So, the concentration of salt in D = $\\dfrac{2b + 7c}{9}$ =\u00a0$\\dfrac{90 – 4c + 7c}{9}$ =\u00a0$\\dfrac{90 + 3c}{9}$<\/p>\n

Required ratio = $\\dfrac{90 + 3c}{9a}$ =\u00a0$\\dfrac{90 + 3c}{9 (30 + c)}$ = $1 : 3$<\/p>\n

Hence, option B is the correct answer.<\/p>\n

8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$
\nas $xy+yz+xz = 0$
\nso equation will be resolved to $x^2 + y^2 + z^2$<\/p>\n

9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

We know that $x^2 – x – 1=0$
\nTherefore $x^4 = (x+1)^2 = x^2+2x+1 = x+1 + 2x+1 = 3x+2$
\nTherefore, $2x^4 = 6x+4$<\/p>\n

We know that $x>0$ therefore, we can calculate the value of $x$ to be $\\frac{1+\\sqrt{5}}{2}$
\nHence, $2x^4 = 6x+4 = 3+3\\sqrt{5}+4 = 3\\sqrt{5}+7$<\/p>\n

10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

We have,\u00a0$\\mid x^2 – x – 6 \\mid = x + 2$<\/p>\n

=> |(x-3)(x+2)|=x+2<\/p>\n

For x<-2, (3-x)(-x-2)=x+2<\/p>\n

=> x-3=1\u00a0 \u00a0=>x=4 (Rejected as x<-2)<\/p>\n

For -2$\\le\\ $x<3,\u00a0(3-x)(x+2)=x+2\u00a0 \u00a0 =>x=2,-2<\/p>\n

For x$\\ge\\ $3,\u00a0(x-3)(x+2)=x+2\u00a0 \u00a0=>x=4<\/p>\n

Hence the product =4*-2*2=-16<\/p>\n

XAT Solved Previous papers<\/a><\/p>
\n

Take XAT Mock Test<\/a><\/p><\/p>\n

We hope this Quant Questions Set-2 PDF for XAT with Solutions will be helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Quant Questions for XAT Set-2 PDF Download important Quant Questions Set-2 for XAT PDF based on previously asked questions in the XAT exam. Practice Quant Questions\u00a0 Set-2 PDF for the XAT exam. Question 1:\u00a0If R = $(30^{65}-29^{65})\/(30^{64}+29^{64})$ ,then a)\u00a0$0<R\\leq0.1$ b)\u00a0$0.1<R\\leq0.5$ c)\u00a0$0.5<R\\leq1.0$ d)\u00a0$R>1.0$ Question 2:\u00a0If x = $(16^3 + 17^3+ 18^3+ 19^3 )$, then x divided […]<\/p>\n","protected":false},"author":42,"featured_media":39765,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3,3167,125,1352,350,362,366],"tags":[3415,1425],"class_list":{"0":"post-39757","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-cat","8":"category-downloads-en","9":"category-featured","10":"category-iift-en","11":"category-iift","12":"category-snap","13":"category-xat","14":"tag-quant-questions","15":"tag-xat"},"better_featured_image":{"id":39765,"alt_text":"Quant Questions for XAT Set-2 PDF","caption":"Quant Questions for XAT Set-2 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