790+ Mocks – Just Rs. 194. Use coupon: SBIDREAM70<\/a><\/p>\n Question 1:\u00a0<\/b>In 8 years, Ramesh will be twice as old as Farheen was 8 years ago. If Ramesh is now 16 years older than Farheen, the present age of Farheen is :<\/p>\n a)\u00a032<\/p>\n b)\u00a040<\/p>\n c)\u00a036<\/p>\n d)\u00a044<\/p>\n e)\u00a024<\/p>\n Question 2:\u00a0<\/b>A was thrice as old as B when C was 2 years old. A will be twice as old as B when C is 10 years old. How old will B be when the difference between the ages of A and B will be equal to the age of C?<\/p>\n a)\u00a016 years<\/p>\n b)\u00a022 years<\/p>\n c)\u00a018 years<\/p>\n d)\u00a024 years<\/p>\n e)\u00a032 years<\/p>\n Question 3:\u00a0<\/b>The ratio of P\u2019s age 3 years ago and Q\u2019s age after 5 years is 6 : 13. If at present, the ratio of their ages is 11 : 20, then what is the ratio of P\u2019s age 2 years hence and Q\u2019s age 5 years ago?<\/p>\n a)\u00a0$\\frac{9}{11}$<\/p>\n b)\u00a0$\\frac{7}{9}$<\/p>\n c)\u00a0$\\frac{5}{13}$<\/p>\n d)\u00a0$\\frac{7}{11}$<\/p>\n e)\u00a0$\\frac{8}{13}$<\/p>\n Question 4:\u00a0<\/b>The sum of the ages of a P and Q is 53 years. Four years ago, the product of their ages was two times the square of Q\u2019s age. The present age of P is?<\/p>\n a)\u00a034 years<\/p>\n b)\u00a038 years<\/p>\n c)\u00a032 years<\/p>\n d)\u00a042 years<\/p>\n e)\u00a036 years<\/p>\n Question 5:\u00a0<\/b>The present age of Sanjay is 40 percent of his mother\u2019s present age. 8 years from now, he will be half as old as his mother. What is the sum of present ages of Sanjay and his mother?<\/p>\n a)\u00a042 years<\/p>\n b)\u00a070 years<\/p>\n c)\u00a056 years<\/p>\n d)\u00a084 years<\/p>\n e)\u00a050 years<\/p>\n Take a Free Mock test for SBI PO<\/a><\/p>\n SBI PO Solved Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>The average marks of a class increased by $\\large\\frac{1}{3}$ when a student\u2019s marks were wrongly entered as 64 instead of 46. The number of students in the class are<\/p>\n a)\u00a054<\/p>\n b)\u00a058<\/p>\n c)\u00a063<\/p>\n d)\u00a057<\/p>\n e)\u00a0None of these<\/p>\n Question 7:\u00a0<\/b>The average age of students in a class is 12. If the average age of boys is 18 and the average age of girls is 14 and If the number of boys are 22, then find the total number of students in the class<\/p>\n a)\u00a052<\/p>\n b)\u00a055<\/p>\n c)\u00a060<\/p>\n d)\u00a058<\/p>\n e)\u00a0None of these<\/p>\n Question 8:\u00a0<\/b>In the first 20 overs of a cricket game, the run rate was 4.5, What should be the run rate in the next 30 overs of the game to chase the target of 210?<\/p>\n a)\u00a06<\/p>\n b)\u00a04<\/p>\n c)\u00a08<\/p>\n d)\u00a06.5<\/p>\n e)\u00a0None of these<\/p>\n Question 9:\u00a0<\/b>In a society, 60% of residents own a car and 40% own a bike. Nobody owns car and bike both. Also, 25% of the residents work in the banking sector. If 20% of those who work in the banking sector own a bike, then the residents who own a car and work in non-banking sector form what percentage of the total people?<\/p>\n a)\u00a025%<\/p>\n b)\u00a040%<\/p>\n c)\u00a037.5%<\/p>\n d)\u00a060%<\/p>\n e)\u00a042.5%<\/p>\n Question 10:\u00a0<\/b>Amit scored 10% less marks than Sumit and 20% more marks than Rakesh. If Tina scored 35 marks more than Amit and Sumit scored 20 marks less than Tina, what is the absolute difference between the marks of Tina and Rakesh?<\/p>\n a)\u00a075<\/p>\n b)\u00a082.5<\/p>\n c)\u00a037.5<\/p>\n d)\u00a057.5<\/p>\n e)\u00a040<\/p>\n Banking Study Material – 18000 Questions<\/a><\/p>\n Daily Free SBI Online Tests<\/a><\/p>\n Instructions<\/b><\/p>\n The following table shows the breakup of the marks scored by students of Class X of Sharda Mandir in Maths for 5 years from 2001-2005.<\/p>\n Note: A student can score a maximum of 100 marks.<\/p>\n Question 11:\u00a0<\/b>The students scoring below 40 are said to have failed the exam whereas the students scoring in between are 60-90 said to have passed the exam with distinction. a)\u00a0420:313<\/p>\n b)\u00a0417:305<\/p>\n c)\u00a0323:251<\/p>\n d)\u00a04:3<\/p>\n e)\u00a05:4<\/p>\n Question 12:\u00a0<\/b>The students scoring between 80 and 90 marks is said to have passed the exam with A grade. How many students got A grade during the given period?<\/p>\n a)\u00a0925<\/p>\n b)\u00a0775<\/p>\n c)\u00a0825<\/p>\n d)\u00a0875<\/p>\n e)\u00a0900<\/p>\n Question 13:\u00a0<\/b>What is the ratio of the number of students scoring marks in between 20 and 40 and the number of students scoring marks in between 80 and 90 in 2004?<\/p>\n a)\u00a01:2<\/p>\n b)\u00a03:4<\/p>\n c)\u00a02:1<\/p>\n d)\u00a04:3<\/p>\n e)\u00a05:3<\/p>\n Question 14:\u00a0<\/b>What is the number of students scoring in between 40 and 60 in 2005?<\/p>\n a)\u00a0450<\/p>\n b)\u00a0470<\/p>\n c)\u00a0460<\/p>\n d)\u00a0480<\/p>\n e)\u00a0440<\/p>\n Question 15:\u00a0<\/b>How many students scored less than 40 in the year 2003?<\/p>\n a)\u00a0550<\/p>\n b)\u00a0475<\/p>\n c)\u00a0450<\/p>\n d)\u00a0425<\/p>\n e)\u00a0600<\/p>\n Question 16:\u00a0<\/b>To meet the demand during summer, a milkman mixed water with pure milk in 2:5. Then he sold the same mixture at 20% more than normal price. By what percent his revenue will increase during summers? (Water is available without any cost)<\/p>\n a)\u00a060 percent<\/p>\n b)\u00a050 percent<\/p>\n c)\u00a040 percent<\/p>\n d)\u00a068 percent<\/p>\n e)\u00a063 percent<\/p>\n Question 17:\u00a0<\/b>A retailer bought 5 dozen notebooks. He got these notebooks at 30% less than printed price. To clear his stock soon he offered 1 notebook free on buying 4 notebooks. If the retailer sold all notebooks on printed price, what\u2019s is retailer\u2019s profit percentage in this transaction?<\/p>\n a)\u00a028.57%<\/p>\n b)\u00a021.35%<\/p>\n c)\u00a012.48%<\/p>\n d)\u00a042.89%<\/p>\n e)\u00a014.28%<\/p>\n Question 18:\u00a0<\/b>Krishna purchased a new laptop. After 2 years, he sold it to Pradeep at a discount of 30% when compared to the actual cost price. Pradeep spent 680 for repairing the laptop after purchase. After using it for 2 years, he sold it to Ravi at a discount of 30% when compared to the total amount he invested in the laptop. If Ravi paid exactly half the amount which Krishna paid, find the amount that Krishna paid when he purchased the laptop?<\/p>\n a)\u00a047600<\/p>\n b)\u00a036600<\/p>\n c)\u00a068000<\/p>\n d)\u00a072000<\/p>\n e)\u00a0None of the above.<\/p>\n Question 19:\u00a0<\/b>Gugan sold a computer to Raj by marking the price up by 20%. Raj then sold the computer to Hari at a profit of 30%. Hari sold the computer to Balu at a profit of 50%. If Balu bought the computer by paying Rs. 1560 more than the price at which Hari bought the computer, what is the actual cost price of the computer?<\/p>\n a)\u00a0Rs.1000<\/p>\n b)\u00a0Rs.2000<\/p>\n c)\u00a0Rs.3000<\/p>\n d)\u00a0Rs.4000<\/p>\n e)\u00a0Rs.5200<\/p>\n Question 20:\u00a0<\/b>A shopkeeper marks up the price of an article by x%. Then, he offers a discount of (x\/2)%.If he ends up with a loss of 48%, then the value of x is<\/p>\n a)\u00a0120<\/p>\n b)\u00a080<\/p>\n c)\u00a0140<\/p>\n d)\u00a0160<\/p>\n e)\u00a060<\/p>\n 200+ Free GK Tests<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let us assume the present age of Farheen = x years Ramesh\u2019s age after 8 years will be = x+16+8 = x+24 years $\\Rightarrow$ x+24 = 2*(x-8) So the present age of Farheen will be 40 years.<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the age of A and B initially be \u2018a\u2019 and \u2018b\u2019 respectively.<\/p>\n When C = 2, a\/b = 3 When C is as old as the difference between the ages of A and B, i.e, when C is 24-8 = 16 years old, age of B will be 8+14 = 22 years. Therefore, option B is the right answer.<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n We have, Hence, option D is the right answer.<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the age of P and Q be p and q respectively. 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the present age of Sanjay\u2019s mother be \u2018x\u2019. So Sanjay\u2019s present age will be .4x General Knowledge Questions & Answers PDF<\/a><\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the number of students be ‘x’<\/p>\n Given that the average is increased by $\\Large\\frac{1}{3}$ due to the increase of $18$ marks ( $64$ $-$ $46$ $=$ $18$ marks )<\/p>\n Actual increase in marks $=$ $\\Large\\frac{1}{3}$ $\\times$ x<\/p>\n $\\Rightarrow$ $\\Large\\frac{1}{3}$ $\\times$ x $=$ $18$<\/p>\n $\\therefore$ Number of students(x) $=$ $54$<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the total age of boys be \u2018B\u2019<\/p>\n Average age of boys = 18<\/p>\n $\\Large\\frac{B}{22}$ $=$ $18$<\/p>\n $\\Rightarrow$ B $=$ $396$<\/p>\n Let the total age of Girls be ‘G’<\/p>\n Let the number of girls be ‘x’<\/p>\n Average age of girls $=$ $14$<\/p>\n $\\Large\\frac{G}{x}$ $=$ $14$<\/p>\n $\\Rightarrow$ G $=$ $14$x<\/p>\n Total average of students $=$ $12$<\/p>\n Total age of students $=$ Total age of boys $+$ Total age of girls $=$ $396$ $+$ $14$x<\/p>\n Total number of students $=$ Number of boys $+$ Number of girls $=$ $22$ $+$ x<\/p>\n Average age of class $=$ $12$<\/p>\n $\\Rightarrow$ $\\Large\\frac{396 + 14x}{22 + x}$ $=$ $12$<\/p>\n $\\Rightarrow$ $396$ $+$ $14$x $=$ $462$ $+$ $12$x<\/p>\n $\\Rightarrow$ $2$x $=$ $66$<\/p>\n $\\Rightarrow$ x $=$ $33$<\/p>\n $\\therefore$ Number of girls $=$ $33$<\/p>\n Total number of students $=$ $22$ $+$ $33$ $=$ $55$<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Run-rate = 4.5 runs per over<\/p>\n $1$ over $\\rightarrow$ $4.5$ runs Total target $=$ $210$ runs<\/p>\n Remaining score $=$ $210$ $-$ $90$ $=$ $120$ runs<\/p>\n Remaining overs $=$ $30$ overs<\/p>\n $\\therefore$ Required run-rate $=$ $\\Large\\frac{120}{30}$ $=$ $4$ runs per over<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the total number of residents be 100x 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let Sumit scored 100x marks 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Number of students scoring in between 0-20 = Total students appeared – Number of students scoring above 20 Thus, the required ratio = 2085:1525 = 417:305 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Number of students scoring in between 0-20 = Total students appeared – Number of students scoring above 20 Thus, 875 students got A grade during the given period. 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Number of students scoring in between 0-20 = Total students appeared – Number of students scoring above 20 Thus, the required ratio = 125:250 = 1:2 14)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Number of students scoring in between 0-20 = Total students appeared – Number of students scoring above 20 Thus, 440 students scored in between 40 and 60 in 2005. 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Number of students scoring in between 0-20 = Total students appeared – Number of students scoring above 20 Thus, 425 students scored less than 40 in the year 2003. 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let us assume that on a normal day he sells milk at x rupee per litre and he sells y litre of milk He mixed water with milk in ratio of 2:5 so net quantity = $\\frac{2+5}{5} * y$ = 1.4y So revenue generated in summers = 1.4y * 1.2x = 1.68xy 17)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Number of notebooks that retailer bought = 5 * 12 = 60<\/p>\n Let us assume that printed price on each notebook = 100<\/p>\n The retailer bought these notebooks at 30% less = 70 Rupees\/notebook<\/p>\n Sum invested by the retailer = 60*70 = 4200<\/p>\n By selling 60 notebooks he will get money for = $\\frac{4}{5}*60$ = 48 notebooks<\/p>\n Money collected by the retailer by selling these notebooks = 48*100 = 4800<\/p>\n Hence, retailer\u2019s profit percentage = $\\frac{4800-4200}{4200}*100$ = 14.28%<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let assume Krishna paid 100x when he purchased the laptop. 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the worth of the computer to be Rs. x. 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the price of the article be $P$. => $x = 160$% and hence, option D is the right answer.<\/p>\n Banking 1 Year Plan – 350 Mocks for just Rs. 499<\/a><\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/di-2409-1-1.PNG\")
<\/figure>\n
\nWhat is the ratio of the number of students that failed and the number of students that got distinction during the given period?<\/p>\n
\nSo present age of Ramesh will be = x + 16 years<\/p>\n
\nFarheen\u2019s age 8 years ago = x – 8 years<\/p>\n
\n$\\Rightarrow$ x= 40 years<\/p>\n
\n=> a = 3b
\nWhen C = 10, 8 years have passed.
\n=> (a+8)\/(b+8) = 2
\na+8 = 2b + 16
\na = 2b+8
\n=> b = 8 years and a = 24 years.<\/p>\n
\n$\\frac{P – 3}{Q + 5} = \\frac{6}{13}$
\nSo, 13P – 39 = 6Q + 30
\n13P – 6Q = 69
\n$\\frac{P}{Q} = \\frac{11}{20}$
\n20P – 11Q = 0
\nSolving both the equations we have,
\nP = 33 years and Q = 60 years
\nSo, $\\frac{P + 2}{Q – 5} = \\frac{35}{55} = \\frac{7}{11}$<\/p>\n
\nWe have, p+q = 53
\n$(p-4)(q-4) = 2 * (q-4)^2$
\np-4 = 2q – 8
\np = 2q – 4
\nSubstituing the value of p in first equation,
\n3q – 4 = 53
\nq = 19
\np = 53 – 19 = 34
\nHence, option A is right choice.<\/p>\n
\nWe have been given that
\n(.4x + 8)*2 = x + 8
\n=> .8x + 16 = x + 8
\n=> .2x = 8
\n=> x = 40
\nHence, the present age of the mother is 40 years. Thus, required sum = 56.<\/p>\n
\n$20$ overs $\\rightarrow$ $20\\times4.5$ $=$ $90$ runs<\/p>\n
\nPeople who own car = 60k
\nPeople who own a bike = 40k
\nPeople who work in the banking sector = 25k
\nPeople in the banking sector who own a bike = 20% of 25k = 5k
\nRest of the people who work in the banking sector must be owning a car
\nNumber of people owning a car and working in banking sector = 25k – 5k = 20k
\nNumber of people owning a car and working in non-banking sector = 60k – 20k = 40k
\nRequired % = 40k\/100k * 100 = 40%
\nHence, option B is the correct answer.<\/p>\n
\nThen, Amit scored 90x marks
\nAmit\u2019s score is 20% more than that of Rakesh.
\nSo, Rakesh\u2019s score = 75x
\nTina\u2019s score = Amit\u2019s score + 35 = 90x + 35…..(i)
\nTina\u2019s score = Sumit\u2019s score + 20 = 100x + 20…(ii)
\nOn comparing (i) and (ii),
\n90x + 35 = 100x + 20
\nOn solving, we get x = 1.5
\nSo, Rakesh\u2019s score = 75x = 75* 1.5 = 112.5
\nAnd Tina\u2019s score = 90x + 35 = 170
\nDifference = (170 – 112.5) = 57.5
\nHence, option D is the correct answer.<\/p>\n
\nNumber of students scoring in between 20-40 = Number of students scoring above 20 – Number of students scoring above 40
\nNumber of students scoring in between 40-60 = Number of students scoring above 40 – Number of students scoring above 60
\nNumber of students scoring in between 60-80 = Number of students scoring above 60 – Number of students scoring above 80
\nNumber of students scoring in between 80-90 = Number of students scoring above 80 – Number of students scoring above 90
\nNumber of students scoring in between 90-100 = Number of students scoring above 90<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/di-2409-1-2.PNG\")
<\/figure>\n
\nHence, option B is the correct answer.<\/p>\n
\nNumber of students scoring in between 20-40 = Number of students scoring above 20 – Number of students scoring above 40
\nNumber of students scoring in between 40-60 = Number of students scoring above 40 – Number of students scoring above 60
\nNumber of students scoring in between 60-80 = Number of students scoring above 60 – Number of students scoring above 80
\nNumber of students scoring in between 80-90 = Number of students scoring above 80 – Number of students scoring above 90
\nNumber of students scoring in between 90-100 = Number of students scoring above 90<\/p>\n<\/figure>\n
\nHence, option D is the correct answer.<\/p>\n
\nNumber of students scoring in between 20-40 = Number of students scoring above 20 – Number of students scoring above 40
\nNumber of students scoring in between 40-60 = Number of students scoring above 40 – Number of students scoring above 60
\nNumber of students scoring in between 60-80 = Number of students scoring above 60 – Number of students scoring above 80
\nNumber of students scoring in between 80-90 = Number of students scoring above 80 – Number of students scoring above 90
\nNumber of students scoring in between 90-100 = Number of students scoring above 90<\/p>\n<\/figure>\n
\nHence, option A is the correct answer.<\/p>\n
\nNumber of students scoring in between 20-40 = Number of students scoring above 20 – Number of students scoring above 40
\nNumber of students scoring in between 40-60 = Number of students scoring above 40 – Number of students scoring above 60
\nNumber of students scoring in between 60-80 = Number of students scoring above 60 – Number of students scoring above 80
\nNumber of students scoring in between 80-90 = Number of students scoring above 80 – Number of students scoring above 90
\nNumber of students scoring in between 90-100 = Number of students scoring above 90<\/p>\n<\/figure>\n
\nHence, option E is the correct answer.<\/p>\n
\nNumber of students scoring in between 20-40 = Number of students scoring above 20 – Number of students scoring above 40
\nNumber of students scoring in between 40-60 = Number of students scoring above 40 – Number of students scoring above 60
\nNumber of students scoring in between 60-80 = Number of students scoring above 60 – Number of students scoring above 80
\nNumber of students scoring in between 80-90 = Number of students scoring above 80 – Number of students scoring above 90
\nNumber of students scoring in between 90-100 = Number of students scoring above 90<\/p>\n<\/figure>\n
\nHence, option D is the correct answer.<\/p>\n
\nSo revenue generated on normal day = x*y = xy<\/p>\n
\nNew selling price during summer = 1.2x rupees per litre<\/p>\n
\nHence we can say that revenue will go up by 68 percent.<\/p>\n
\nSo Krishna sold it to Pradeep for = $\\frac{100-30}{100}*100x$ = 70x
\nAnd Pradeep also spent 680 rupees for repairing
\nSo total amount invested by Pradeep = 70x+680
\nSo the amount Ravi paid to Pradeep = $\\frac{100-30}{100}*(70x+680)$ = 49x+476
\nWe also know that Ravi paid exactly half of what Krishna paid
\ni.e. 0.5*(100x) = 49x+476
\n$\\Rightarrow$ x = 476
\nSo the amount for which Krishna purchased the laptop = 100*476 = 47600<\/p>\n
\nLet the price at which Raj bought the computer be 1.2x.
\nPrice at which Hari bought = 1.3*1.2x = 1.56x
\nPrice at which Balu bought = 1.5*1.56x = 2.34x
\nDifference between the prices at which Balu and Hari bought = 2.34x – 1.56x = 0.78x
\nIt has been given that 0.78x = 1560
\nx = Rs. 2000.
\nTherefore, option B is the right answer.<\/p>\n
\nLet us consider $x$ in decimal form for the ease of calculation.
\n$(1+x)(1-x\/2)P = 0.52P$
\n$(1+x)(2-x) = 1.04$
\n$2-x+2x -x^2 = 1.04$
\n$-x^2+x = -0.96$
\n$x^2-x-0.96 = 0$
\n$x = 1.6$ or $x = -0.6$
\nWe can ignore the negative root since it has been given that the shopkeeper marks up the price of the product.<\/p>\n