Cambridge IGCSE Mathematics - International 0607 — 2022 May/June Paper 4 · Variant 1

1 y 3 2 1 – 4 – 3 – 2 – 1 0 1 2 3 4 x T – 1 – 2 P – 3 - 2 (a) Translate triangle T by the vector [2] e 2o. (b) Reflect triangle T in the line y = 0 .5 . [2] (c) Describe fully the single transformation that maps triangle P onto triangle T. ..................................................................................................................................................... ..................................................................................................................................................... [3] (d) Enlarge triangle P with scale factor - 2 , centre ( 3, - 1) . [2]

Mark scheme: Question Answer Marks Partial Marks 1(a) correct triangle 2 k  2  (−4, 1) (−3, 1) (−3, 1.5) B1 for translation  or   or 2  k   1  translation through    1  1(b) correct triangle 2 B1 for reflection in either y = k or (−1, 1.5) (−1, 2) (−2, 2) x = 0.5 1(c) Rotation 3 B1 for each 180⁰ (1, −1.5) 1(d) correct triangle 2 B1 for any enlargement scale factor −2 (3, 1) (3, 2) (1, 1)

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2 (a) The cumulative frequency curve shows the marks for 300 students in a history test. 300 250 200 Cumulative 150 frequency 100 50 0 0 10 20 30 40 50 History mark (i) Find an estimate for the median. ................................................. [1] (ii) Estimate the number of students with a mark of more than 20. ................................................. [2] (iii) 70% of the students pass the test. Find the pass mark. ................................................. [2] (b) The table shows the marks for 100 students in a geography test. Mark m 10 1 m G 20 20 1 m G 30 30 1 m G 40 40 1 m G 50 Frequency 2 28 57 13 Calculate an estimate of the mean. ................................................. [2] (c) The table shows the marks for 9 students in chemistry and in physics. Chemistry 33 28 39 40 22 25 38 43 36mark (x) Physics 45 32 26 49 18 36 29 40 35mark (y) (i) Find the equation of the regression line for y in terms of x. y = ................................................ [2] (ii) What type of correlation is seen in this data? ................................................. [1] (iii) Use your answer to part (c)(i) to estimate the physics mark for a student with a mark of 30 in chemistry. ................................................. [1]

Mark scheme: 2(a)(i) 31 1 2(a)(ii) 260 2 B1 for 40 seen 2(a)(iii) 27 2 70 30 M1 for  300 soi or  300 100 100 2(b) 33.1 2 M1 for at least three mid-values soi 2(c)(i) y  0.618 x  13.6 2 B1 for 0.618x + k or kx + 13.6 or 0.62x + 14 2(c)(ii) positive 1 2(c)(iii) 32 or 32.0 to 32.6 1 FT their (c)(i) if linear eqn

More questions on Cumulative frequency diagrams

3 y 5 – 3 0 1 x – 5 1 f ( x) = 2 x + 4 - 2 x (a) On the diagram, sketch the graph of y = f ( x) for values of x between - 3 and 1. [3] (b) Write down the equation of the asymptote of the graph. ................................................. [1] (c) Find the coordinates of the local maximum. (....................... , .......................) [1] (d) g( )x = x 3 - 5x for - 3 G x G 1. Solve f ( x) G g( x) . ................................................. [4]

Mark scheme: 3(a) correct sketch 3 B2 for correct branches but joined or touching y-axis B1 for one correct branch 3(b) x = 0 1 3(c) (−1, 1) 1 3(d) −2.31 ⩽ x < 0 and 0 < x ⩽ 0.388 4 B3 for −2.3 ⩽ x ⩽ 0.388 or −2.31 ⩽ x ⩽ 0.39 or B2 for –2.3 ⩽ x ⩽ 0.39 or –2.31 ⩽ x or x ⩽ 0.388 or B1 for – 2.31 or 0.388 seen or for correct sketch

More questions on Inequalities

4 (a) $216 is shared in the ratio 5 : 1. Work out the larger share. $ ................................................ [2] (b) Luis shares some money between Ali, Betty and Clare in the ratio 3 : 4 : 6. Ali receives $171. Find the total amount of money Luis shared. $ ................................................ [2] (c) Farima invests $1400 in a savings account paying simple interest at a rate of 2.5% per year. Calculate the total amount in the account at the end of 3 years. $ ................................................ [3] (d) Emir invests $3000 at a rate of 2% per year compound interest. (i) Calculate the value of Emir’s investment at the end of 4 years. $ ................................................. [2] (ii) Find the number of complete years until Emir’s investment is first worth more than $4000. ................................................. [4]

Mark scheme: 4(a) 180 2 216 M1 for [ 5] 5  1 4(b) 741 2 171 M1 for 3 4(c) 1505 3 2.5 M2 for 1400 + 3(1400 × ) oe 100 2.5 or M1 for 3(1400 × ) oe 100 4(d)(i) 3247.30 or 3250 2 4  2  M1 for 3000  1  oe    100  4(d)(ii) 15 4 B3 for 14.5 or 14.52 to 14.53 seen  4000  or M3 for n log(1.02)  log   oe  3000  or for correct trials reaching 14 and 15 or good sketch indicating value between 14 and 15 4000 or M2 for 1.02n = 3000 or at least three correct trials for n > 4 or suitable graph or M1 for 3000 × 1.02n = 4000 soi by at least 2 correct trials for n > 4

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5 A sequence of patterns is made using grey tiles and white tiles. Pattern 1 Pattern 2 Pattern 3 (a) Complete the table. Pattern number 1 2 3 4 n Number of grey tiles 6 10 Number of white tiles 0 2 [6] (b) Find and simplify an expression for the total number of tiles in Pattern n. ................................................. [1] (c) Pattern k has a total of 600 tiles. Find the number of grey tiles in Pattern k. ................................................. [4] (d) The tiles in a pattern are put in a bag. 5 The probability of taking a grey tile from the bag at random is . 12 A tile is taken from the bag at random and replaced. This is repeated 3 times. Find the probability that all 3 tiles are white. ................................................. [2] (e) All the grey tiles from Pattern 4 are put in a bag. Two tiles are taken from the bag at random without replacement. Find the probability that one tile came from a corner of the pattern and the other did not. ................................................. [3]

Mark scheme: 5(a) 6 B2 for all four numbers correct 14 18 4n + 2 oe or B1 for at least two correct B2 for 4 n  2 oe 6 12 n2 – n oe or M1 for 4 n  k B2 for n 2  n oe or M1 for any quadratic or for second differences of 2 seen 5(b) n2 + 3n + 2 or (n + 1)(n + 2) 1 FT their grey + white if both in terms of n 5(c) 94 4 M1 for their k 2  3k  2 = 600 oe M1 for correct method for solving their quadratic M1 for substituting their positive integer (from a quadratic) k into their 4n+2 5(d) 343 2  5  3 oe M1 for 1  oe 1728    12  5(e) 56 3 FT their 18 from (a) for M marks only oe 153 4 14 14 4 M2 for    oe 18 17 18 17 M1 for one product

More questions on Sequences

6 (a) NOT TO C SCALE 5 cm B 30° O A The diagram shows a circle, centre O, with radius 5 cm. BA and BC are tangents to the circle at A and C. Angle ABC = 30° . Calculate the area of the shaded minor segment. ......................................... cm2 [4] (b) O NOT TO height O SCALE 40° 12 cm D E D E The circle, centre O, has radius 12 cm. Angle DOE = 40° . The minor sector DOE is removed. The major sector is formed into a cone by joining OD to OE. Calculate the height of the cone. ............................................ cm [5]

Mark scheme: 6(a) 26.5 or 26.47 to 26.48 4 B1 for AOC = 150 or BOC = 75 soi their150  5 2 M1 for or 360 their 75  5 2  2 360 1 M1 for 5 5 sin[their150] or 2 1 5 5 sin[their 75  2] 2 6(b) 5.5[0] or 5.49 to 5.51... 5 (360  40) M2 for 2 12 = 2r 360 (360  40) 2 oe or  12 = πr12 oe 360 (360  40) or M1 for 2 12 or 360 (360  40) 2  12 360 4 If minor sector SC1 for radius = oe 3 AND M2 for 12 2 their ( radius ) 2 oe dependent on at least M1 or SC1 or M1 for h 2  their ( radius ) 2  12 2 oe dependent on at least M1 or SC1

More questions on Surface area and volume

7 Abbi makes wooden boards in three sizes, small, medium and large. They are all cuboids. The medium board has height 2 cm, width 23 cm and length 50 cm. (a) Calculate the volume of the medium board. ......................................... cm3 [2] (b) The small board is mathematically similar to the large board. The small board has a volume of 287.5 cm 3and a height of 1.15 cm. The large board has a volume of 18400 cm 3. (i) Find the height of the large board. ............................................ cm [3] (ii) Is the medium board mathematically similar to the large board? Explain how you decide. ........................... because ................................................................................................... ............................................................................................................................................. [3]

Mark scheme: 7(a) 2300 2 M1 for 2 × 23 × 50 oe 7(b)(i) 4.6 3 18400 M2 for 1.15  3 oe 287.5 h 3 18400 18400 or M1 for  or 3 1.153 287.5 287.5 287.5 or 3 seen 18400 7(b)(ii) 2 : their (b)(i) or their (a) : 18400 soi M1 FT their figures showing comparison of length ratio or M1 volume ratio not similar A1 Dep on M1M1

More questions on Similarity

8 (a) A is the point ( - 11, 7) and B is the point ( 8 , - 13) . Find the length of AB. ................................................. [3] (b) P is the point ( 2, - 5) and Q is the point ( 6, 11) . Line L is perpendicular to PQ and crosses PQ at point R. The ratio PR : RQ = 3 : 1. Find the equation of line L. ................................................. [6]

Mark scheme: 8(a) 27.6 or 27.58 to 27.59 3 M2 for (( 11)  8) 2  (7 ( 13)) 2 oe or M1 for (( 11)  8) or (7 ( 13)) oe 8(b) y  14 x  8 14 oe 6 B5 for answer  14 x  8 14 OR B2 for (5, 7) or B1 for (5, k) or (k, 7) 11 5 M1 for oe (=m1) 6  2 1 M1 for grad = their m1 M1 for substituting their (5, 7) into y = (their m)x + c

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9 (a) f ( x) = 2 x + 3 g ( x) = 2 - 4 x h ( x) = 3x (i) Find f(5). ................................................. [1] (ii) Find and simplify g(f(x)). ................................................. [2] (iii) Find g -1 ( x) . g -1 ( x) = ................................................ [2] (iv) Solve h ( x) = 48 . ................................................. [2] (b) (i) The diagram shows a sketch of the graph of y = j ( x) . y 10 j(x) 5 – 10 – 5 0 5 10 x – 5 – 10 On the same diagram, sketch the graph of y = j ( x + 2) . [1] (ii) The diagram shows the graphs of y = k ( x) and y = m ( x) . y 10 k(x) m(x) 5 0 x – 5 3 – 5 – 10 Write k ( x) in terms of m ( x) . k ( x) = ................................................ [1]

Mark scheme: 9(a)(i) 13 1 9(a)(ii) 8 x  10 oe final answer 2 M1 for 2  4(2 x  3) or better seen 9(a)(iii) 2 x 2 M1 for correct first step oe final answer 4 y 1 y  4 x  2 or   x or x  2  4 y 4 2 y – 2 = – 4x 9(a)(iv) 3.52 or 3.523 to 3.524 or log 3 48 or 2 M1 for log48  log3 x or better log48 or suitable sketch log3 9(b)(i) 1 9(b)(ii) 2m(x) 1

More questions on Functions

10 (a) Simplify fully. 4 x 2 y x ' 3 12 y ................................................. [2] (b) Write as a single fraction in its simplest form. 1 x - 3 - x - 3 2 ................................................. [3] (c) The nth term of a sequence is an 2 + bn - 5 . The second term of this sequence is - 3 and the third term is 4. Find the value of a and the value of b. You must show all your working. a = ................................................ b = ................................................ [6]

Mark scheme: 10(a) 16xy 2 Final answer 2 4 x 2 y 12 y M1 for  or better 3 x 10(b)  x 2  6 x  7 3 B1 for 2[ 1]  ( x  3)( x  3) Final answer B1 for 2( x  3) as denominator 2( x  3) 10(c) 2 2  a  2  b  5 3 oe M2 M1 for either 32  a 3 b  5  4 oe correctly equating one set of coefficients M1 FT or making a or b subject of one equation correct method for eliminating one M1 FT variable or correctly substituting in other equation [a=] 2 B2 B1 for each [b=] −3

More questions on Equations

11 NOT TO SCALE 2.1 m 110° 110° 0.9 m The diagram shows the symmetrical cross-section of a ditch containing water. The angle between the base and each side of the ditch is 110°. The width of the base is 0.9 m and the depth of the water is 2.1 m. The ditch is 100 m long. (a) Calculate the volume of water in the ditch. .............................................m3 [4] (b) On a different day, the ditch contains 300 m 3of water. Water is pumped out of the ditch at a rate of 4.2 litres per second. Calculate the time taken to empty the ditch completely. Give your answer in hours and minutes, correct to the nearest minute. ................... h ................... min [4]

Mark scheme: 11(a) 349 or 350 or 349.4 to 349.5… 4 M3 for [0.9  2.1  2  12 (2.1  2.1tan 20)] × 100 oe 1 or (0.9  (2  2.1tan 20  0.9))  2.1 × 2 100 oe or M2 for area of cross section 0.9  2.1  2[ 12 (2.1  2.1tan 20)] oe 1 or (0.9  (2  2.1tan 20  0.9))  2.1 2 oe or M1 for tan 20 x 2.1 If 0 scored SC1 for triangle marked/drawn with 20 or 70 11(b) 19 h 50 (or 51) min 4 B3 for 19.8 or 19.84… 300  1000 or M2 for oe 4.2  60  60 or M1 for 300  1000 or 4.2  60  60 or for their volume divided by their rate

More questions on Surface area and volume

12 B NOT TO A SCALE 80° 9 m 16 m D 115° 22 m C (a) Calculate the area of triangle BCD. ............................................ m2 [2] (b) Calculate angle ADB. Angle ADB = ................................................ [6]

Mark scheme: 12(a) 160 or 159.5… 2 1 M1 for  16  22  sin115 oe 2 12(b) 84[.0] or 84.02 to 84.03 6 B3 for 32.2 or 32.21.. soi OR M2 for 16 2  22 2 2 16  22  cos115 or M1 for 16 2  22 2  2  16  22  cos115 AND 9sin80 M2 for dependent on their (32.21) cosine rule used sin ABD sin80 or M1 for  9 their (32.21)

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